The acceleration of the ball while the bat is in contact with it is 400 m/s².
To find the acceleration of the ball while the bat is in contact with it, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
Given:
Force applied by the bat (F) = 100 N
Frictional force opposing motion (f) = 20 N
Mass of the ball (m) = 0.2 kg
Net force acting on the ball can be calculated as:
Net force (F_net) = F - f
Substituting the given values:
F_net = 100 N - 20 N
F_net = 80 N
Using Newton's second law:
F_net = ma
Solving for acceleration (a):
a = F_net / m
a = 80 N / 0.2 kg
a = 400 m/s²
Therefore, the acceleration of the ball while the bat is in contact with it is 400 m/s².
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4. A water reservoir receives water from approximately 100 hectares of land.
a) If the average annual rainfall is 95-cm per year how much water (m3) will the reservoir collect per year?
b) If you need to pump water from the reservoir at 4-gpm through a 8000-ft, 3-inch PVC pipeline to a school that is located at an elevation of 120-ft above the reservoir, what should the rated head of the pump be? Assume the water pressure at the school needs to be 30 psi.
(a). The water reservoir will collect 950,000 m³ of water per year.
(b). The rated head of the pump should be 58.88 m.
(a). To find: How much water (m³) will the reservoir collect per year?
As per data,
Area of land from where water is collected = 100 hectares
Average annual rainfall = 95 cm per year
1 hectare = 10,000 m²
Area of land from where water is collected = 100 hectares
Therefore,
Area = 100 × 10,000
= 1,000,000 m².
Volume of water collected in the reservoir per year = Area × Average annual rainfall
= 1,000,000 m² × 95 cm
= 1,000,000 m² × 0.95 m
= 950,000 m³
Hence, the volume of water is 950,000 m³ of water per year.
(b). To find: The rated head of the pump
As per data,
Flow rate of water = 4 gpm
Length of PVC pipeline = 8000 ft or 2438.4 m
Diameter of PVC pipeline = 3 inch or 0.0762 m
Elevation of school = 120 ft or 36.576 m
Water pressure at the school = 30 psi
Convert flow rate from gallons per minute (gpm) to cubic meter per second (m³/s).
1 US gallon = 0.00378541 m³
Flow rate of water = 4 gpm
= 4 × 0.00378541 m³/s
= 0.01514 m³/s.
Cross-sectional area of 3-inch PVC pipe,
A = π/4 × (0.0762 m)²
= 0.00457 m².
Velocity of water in the PVC pipe,
v = Q/A
Where,
Q = flow rate of water= 0.01514 m³/s
A = cross-sectional area of the pipe= 0.00457 m²
∴ v = 0.01514/0.00457
= 3.31 m/s.
Head loss in PVC pipe,
Hloss = 0.9 × (2438.4/1000) × (v²/2g)
Where,
g = acceleration due to gravity= 9.81 m/s²
∴ Hloss = 0.9 × 2.4384 × (3.31)2/(2 × 9.81)
= 1.62 m.
Total dynamic head (TDH) = Elevation head + friction loss + pressure head
Where,
Elevation head = elevation difference between the reservoir and school
= 120 - 0
= 120 ft
= 36.576 m.
Friction loss = Head loss in PVC pipe = 1.62 m
Pressure head = (30 psi × 0.06895 kg/m³) / g
= 20.685 m
TDH = 36.576 + 1.62 + 20.685
= 58.88 m
Rated head of the pump = TDH = 58.88 m.
Hence, the rated head of the pump is 58.88 m.
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A narrow beam of protons of speed 2000 m/s passes through a uniform 0.1 Tesla magnetic field with the direction of the magnetic field at a 30 degrees angle to the beam. In addition a uniform electric field is present which is perpendicular to both the direction of the beam and the magnetic filed. Calculate the strength of the electric field E in V/m which is needed to keep the beam on a straight line path.
To keep the beam of protons on a straight-line path, we need to balance the magnetic force experienced by the protons with the electric force. The magnetic force experienced by a moving charged particle in a magnetic field is given by the equation:
F_magnetic = q * v * B * sin(theta)
F_magnetic is the magnetic force,
q is the charge of the particle (in this case, the charge of a proton, which is 1.602 x 10^-19 C),
F_electric = q * E
F_magnetic = F_electric
q * v * B * sin(theta) = q * E
E = v * B * sin(theta)
E = 2000 m/s * 0.1 Tesla * sin(30 degrees)
Using the trigonometric identity sin(30 degrees) = 0.5:
E = 2000 m/s * 0.1 Tesla * 0.5
E = 100 V/m
Therefore, the strength of the electric field (E) needed to keep the beam on a straight-line path is 100 V/m.
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The work done by the force on a spring is -0.078 J to compress it about 2.5 cm. Which is the
spring constant? (W = − -1/2kx^2)
According to the question the spring constant is approximately -250 N/m.
To find the spring constant, we'll use the work-energy principle and the equation for work done by a spring.
Given:
Work done (W) = -0.078 J (negative sign indicates work done on the spring)
Compression distance (x) = 2.5 cm = 0.025 m
The equation for work done by a spring is given by:
[tex]\[ W = -\frac{1}{2} kx^2 \][/tex]
Substituting the given values into the equation, we have:
[tex]\[ -0.078 = -\frac{1}{2} k (0.025)^2 \][/tex]
Simplifying the equation, we can solve for the spring constant (k):
[tex]\[ k = \frac{2W}{x^2} \][/tex]
Substituting the given values, we get:
[tex]\[ k = \frac{2(-0.078)}{(0.025)^2} \][/tex]
Calculating the value of k:
[tex]\[ k = \frac{-0.156}{0.000625} \approx -250 \, \text{N/m} \][/tex]
Therefore, the spring constant is approximately -250 N/m.
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Steam initially at a pressure of 700 kN/m² with a dryness fraction of 0.65 passes through a reversible process which appears as a semicircle when plotted on a temperature entropy chart. The final pressure is 100 kN/m2 and the final temperature is the same as the initial steam temperature at the pressure of 700 kN/m². During the process the maximum temperature is 230 °C. For this process, determine (a) the heat transfer per kilogram of steam (b) the work done per kilogram of steam. f(a) 1282 kJ/kg; (b) 591.7 kJ/kg] Steam at a pressure of 2 MN/m' and temperature 250C is expanded to a pressure of 0.32 MN/m2 according to the law PV125 = constant. For this expansion, determine (a) the final condition of the steam (b) the specific heat transfer the change of specific entropy [(a) 0.847; (b) – 163.9 kJ/kg: (C) -0.383 kJ/kg K] Steam at a pressure of 1.9 MN/m² and with a temperature of 225 °C is expanded isentropically to a pressure of 0.3 MN/mº. It is then further expanded hyperbolically to a pressure of 0.12 MN/m². Using steam tables, determine (a) the final condition of the steam (b) the change of specific entropy during the hyperbolic process [(a) 0.954; (b) 0.572 7 kJ/kg K. an increase)
According to the question (a) [tex]\( R = 4.13 \, \text{kJ/(kgK)} \)[/tex] , (b) [tex]\( V = 6.84 \, \text{m³/kg} \) at \( T = 70°C \)[/tex] and [tex]\( P = 2.07 \, \text{bar} \)[/tex] , (c) change of specific entropy [tex]\( \Delta Q = 304 \, \text{kJ} \)[/tex] , (d) [tex]\( \Delta U = 304 \, \text{kJ} \)[/tex] , (e) [tex]\( \Delta W = 88.3 \, \text{kJ} \)[/tex].
Given:
Density of the gas, [tex]\( \rho = 0.09 \, \text{kg/m³} \) (at \( T = 0°C \) and \( P = 1.013 \, \text{bar} \))[/tex]
Initial volume, [tex]\( V_1 = 5.6 \, \text{m³} \)[/tex]
Initial pressure, [tex]\( P_1 = 1.02 \, \text{bar} \)[/tex]
Initial temperature, [tex]\( T_1 = 0\°C \)[/tex]
Final temperature, [tex]\( T_2 = 50\°C \)[/tex]
Specific heat capacity at constant pressure, [tex]\( c_p = 10.08 \, \text{kJ/(kg\·K)} \)[/tex]
(a) To determine the characteristic gas constant, we can use the ideal gas law:
[tex]\[ P \cdot V = m \cdot R \cdot T \][/tex]
where:
[tex]\( P \)[/tex] is the pressure (in Pa),
[tex]\( V \)[/tex] is the volume (in m³),
[tex]\( m \)[/tex] is the mass of the gas (in kg),
[tex]\( R \)[/tex] is the characteristic gas constant (in J/(kg·K)),
[tex]\( T \)[/tex] is the temperature (in K).
We can rearrange the equation to solve for the characteristic gas constant:
[tex]\[ R = \frac{{P \cdot V}}{{m \cdot T}} \][/tex]
Given that the gas density is [tex]\( \rho = \frac{{m}}{{V}} \)[/tex], we can substitute this into the equation:
[tex]\[ R = \frac{{P}}{{\rho \cdot T}} \][/tex]
Substituting the values:
[tex]\[ R = \frac{{1.013 \times 10^5 \, \text{Pa}}}{{0.09 \, \text{kg/m³} \times (273.15 \, \text{K} + 0)}} \][/tex]
Simplifying:
[tex]\[ R = 4.13 \, \text{kJ/(kg·K)} \][/tex]
Therefore, the characteristic gas constant is [tex]\( R = 4.13 \, \text{kJ/(kg\·K)} \).[/tex]
(b) To calculate the specific volume of the gas at [tex]\( T = 70\°C \) and \( P = 2.07 \, \text{bar} \)[/tex], we can use the ideal gas law:
[tex]\[ P \cdot V = m \cdot R \cdot T \][/tex]
We rearrange the equation to solve for the specific volume:
[tex]\[ V = \frac{{m \cdot R \cdot T}}{{P}} \][/tex]
Substituting the values:
[tex]\[ V = \frac{{0.09 \, \text{kg} \times 4.13 \, \text{kJ/(kg·K)} \times (273.15 \, \text{K} + 70)}}{{2.07 \times 10^5 \, \text{Pa}}} \][/tex]
Simplifying:
[tex]\[ V = 6.84 \, \text{m³/kg} \][/tex]
Therefore, the specific volume of the gas at [tex]\( T = 70\°C \) and \( P = 2.07 \, \text{bar} \) is \( V = 6.84 \, \text{m³/kg} \).[/tex]
(c) To determine the heat transfer during the process where the gas is heated at constant pressure, we can use the first law of thermodynamics:
[tex]\[ \Delta Q = \Delta U + \Delta W \][/tex]
where:
[tex]\( \Delta Q \)[/tex] is the heat transfer (in J),
[tex]\( \Delta U \)[/tex] is the change in internal energy of the gas (in J),
[tex]\( \Delta W \)[/tex] is the work transfer (in J).
At constant pressure, the heat transfer is given by:
[tex]\[ \Delta Q = m \cdot c_p \cdot \Delta T \][/tex]
Substituting the values:
[tex]\[ \Delta Q = 0.09 \, \text{kg} \times 10.08 \, \text{kJ/(kg·K)} \times (50°C - 0°C) \][/tex]
Simplifying:
[tex]\[ \Delta Q = 304 \, \text{kJ} \][/tex]
Therefore, the heat transfer during the process is [tex]\( \Delta Q = 304 \, \text{kJ} \).[/tex]
(d) The change in internal energy of the gas can be calculated using the specific heat capacity at constant pressure:
[tex]\[ \Delta U = m \cdot c_p \cdot \Delta T \][/tex]
Substituting the values:
[tex]\[ \Delta U = 0.09 \, \text{kg} \times 10.08 \, \text{kJ/(kg·K)} \times (50°C - 0°C) \][/tex]
Simplifying:
[tex]\[ \Delta U = 304 \, \text{kJ} \][/tex]
Therefore, the change in internal energy of the gas is [tex]\( \Delta U = 304 \, \text{kJ} \).[/tex]
(e) The work transfer can be calculated using the equation:
[tex]\[ \Delta W = P \cdot \Delta V \][/tex]
where [tex]\( \Delta V \)[/tex] is the change in volume.
Since the process occurs at constant pressure, we can express [tex]\( \Delta V \)[/tex] as:
[tex]\[ \Delta V = V_2 - V_1 \][/tex]
Substituting the values:
[tex]\[ \Delta V = V_2 - V_1 = 6.84 \, \text{m³/kg} - 5.6 \, \text{m³/kg} \][/tex]
Simplifying:
[tex]\[ \Delta V = 1.24 \, \text{m³/kg} \][/tex]
Now, we can calculate the work transfer:
[tex]\[ \Delta W = P \cdot \Delta V = 1.02 \times 10^5 \, \text{Pa} \times 1.24 \, \text{m³/kg} \][/tex]
Simplifying:
[tex]\[ \Delta W = 88.3 \, \text{kJ} \][/tex]
Therefore, the work transfer during the process is [tex]\( \Delta W = 88.3 \, \text{kJ} \).[/tex]
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You walk in a straight line for 78 m at an angle of 158
∘
above the positive x axis Part A What is the x component of your displacement?
The distance traveled by a person is 78 m, and the angle made by the path of the person with the positive x-axis is 158 degrees. To find: The x-component of displacement of the person.
We have the distance traveled by a person, let's call it 'd'.d = 78 m. Next, we will find the x-component of displacement using the given angle. We know that:x-component = d * cos(angle)We need to convert the angle to radians because the cos function expects an angle in radians. So, let's convert the given angle to radians.1 degree = π/180 radians.So, 158 degrees = 158 * π/180 radians= (79π/90) radians. Now, let's substitute the values in the formula of the x-component of displacement:
x-component = d * cos(angle)x-component = 78 * cos(79π/90)x-component = 78 * (-0.3420)x-component = -26.676 m
Therefore, the x-component of displacement is -26.676 m.
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A horizontal massless rigid bar of length L has a pivot at it's mid point and two springs of stiffness k attached at each end. When the bar is rotated through a small angle about the pivot the two springs resist the displacement. What is the torsional stiffness of the system.
The torsional stiffness of the system is equal to the product of the spring stiffness [tex](\(k\))[/tex] and the length of the bar [tex](\(L\))[/tex].
To find the torsional stiffness of the system, we need to consider the restoring torque exerted by the springs when the bar is rotated.
Let's assume the angle of rotation from the equilibrium position is [tex]\(\theta\)[/tex]. The displacement of each end of the bar from its equilibrium position is [tex]\(\frac{L\theta}{2}\).[/tex]
The torque exerted by each spring is given by Hooke's Law for torsion, which states that the torque is proportional to the angular displacement:
[tex]\(\tau = -k\theta\)[/tex]
The negative sign indicates that the torque acts in the opposite direction to the displacement.
Now, let's consider one of the springs. The lever arm for the torque is [tex]\(\frac{L}{2}\),[/tex] and the torque is [tex]\(-k\left(\frac{L\theta}{2}\right) = -\frac{kL\theta}{2}\)[/tex].
Since there are two identical springs attached to the ends of the bar, the total torque exerted by both springs is:
[tex]\(\tau_{\text{total}} = -2\left(\frac{kL\theta}{2}\right) = -kL\theta\)[/tex]
This torque is proportional to the angle of rotation [tex]\(\theta\)[/tex], so we can define the torsional stiffness [tex](\(k_t\))[/tex] as the proportionality constant:
[tex]\(\tau_{\text{total}} = -k_t\theta\)[/tex]
Comparing this with the equation for torque, we can see that the torsional stiffness of the system is:
[tex]\(k_t = kL\)[/tex]
Therefore, the torsional stiffness of the system is equal to the product of the spring stiffness [tex](\(k\))[/tex] and the length of the bar [tex](\(L\))[/tex].
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A car moving to the right (vx >0) starts breaking with constant acceleration a x= −a<0. Due to the acceleration of the car, a ball with mass m attached to a string hanging from the rear-view mirror makes an angle θ with respect to the vertical. (a) Make a free body diagram showing the angle θ. Is the ball to the right or the the left of its original equilibrium position when the car was moving at a constant speed? Explain. (b) Find the angle θ in terms of m,a, and g, or a subset of these quantities. (c) What is the tension in the string? (d) The driver of the car sees that the ball is not moving as the car slows down and incorrectly assumes that it is in equilibrium. According to the driver, what are the magnitude and direction of the force applied to the ball that would keep it in equilibrium? (e) You ask the driver who/what is exerting that force. Can they give you a satisfying answer? Explain.
When a car with constant acceleration brakes, a ball attached to a string hanging from the rear-view mirror makes an angle θ with respect to the vertical.
The angle θ can be found in terms of m, a, and g by considering the equilibrium condition of the horizontal forces.
The tension in the string can be determined by balancing the vertical forces acting on the ball.
The driver incorrectly assumes that the ball is in equilibrium when the car slows down, but in reality, an unbalanced force is required to maintain equilibrium.
The driver cannot provide a satisfying answer to who or what is exerting that force.
(a) In the free body diagram, the ball experiences the force of gravity acting downwards (mg) and the tension in the string pulling upwards (T). The angle θ is formed between the string and the vertical direction.
When the car with constant acceleration brakes, the acceleration acts in the opposite direction to the original motion, causing the ball to swing to the left of its original equilibrium position. This is because the acceleration provides an additional force component that contributes to the swinging motion.
(b) To find the angle θ, we can consider the equilibrium condition of the horizontal forces. The horizontal force acting on the ball is the tension in the string, T, which is balanced by the component of the gravitational force in the horizontal direction. The horizontal component of the gravitational force is mg sinθ.
Therefore, T = mg sinθ. Solving for θ, we have θ = sin^(-1)(T / mg).
(c) The tension in the string can be determined by balancing the vertical forces acting on the ball. The vertical force acting on the ball is the sum of the gravitational force (mg) and the vertical component of the acceleration force (ma).
Since the ball is not accelerating vertically, the sum of these forces must be zero: T - mg cosθ - ma = 0. Solving for T, we have T = mg cosθ + ma.
(d) The driver incorrectly assumes that the ball is in equilibrium when the car slows down. According to the driver's assumption, the magnitude and direction of the force applied to the ball that would keep it in equilibrium would be zero.
However, in reality, an unbalanced force is required to maintain equilibrium, as the ball is not at rest and is experiencing a non-zero acceleration due to the car braking.
(e) When asked who or what is exerting the force to maintain equilibrium, the driver cannot provide a satisfying answer. The force required to maintain equilibrium is the tension in the string, which is provided by the string itself.
The tension arises due to the gravitational force acting on the ball and the acceleration of the car. The driver may not be aware of the specific mechanisms and forces involved, leading to an unsatisfactory explanation.
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The position of a plane is given by the function x = 4.00 t2 − 5.00 t + 5.00, where t is in seconds.
At what time is the velocity of the plane zero?
a) 1.55 s
b) 0.406 s
c)0 s
d)0.625 s
The time at which the velocity of the plane is zero is approximately 0.625 seconds.
The velocity of the plane can be determined by taking the derivative of the position function with respect to time.
Given:
Position function: x = 4.00 t^2 - 5.00 t + 5.00
To find the time at which the velocity is zero, we need to find the time when the derivative of the position function is zero.
Taking the derivative of the position function, we get:
v = d/dt (4.00 t^2 - 5.00 t + 5.00) = 8.00 t - 5.00
Setting the velocity equal to zero and solving for t:
8.00 t - 5.00 = 0
8.00 t = 5.00
t = 5.00 / 8.00
t ≈ 0.625 s
Therefore, the time at which the velocity of the plane is zero is approximately 0.625 seconds.
Hence, the correct answer is option d) 0.625 s.
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A 50−g mass, attached to a 0.10−N/m spring fixed at the other end, oscillates on a smooth, horizontal surface. At time t=0.50 s, the mass is 14 cm to the right of the centre of oscillation O and is travelling away from O at 23 cm/s. Determine the position-time x(t) equation for this shm (take right to be the direction of positive x ).
The position-time equation for the simple harmonic motion (SHM) is:
x(t) = 0.14 * cos(√2t + 1.446),
where x(t) is the position of the mass at time t, and 0.14 is the amplitude of the motion.
To determine the position-time equation for the simple harmonic motion (SHM) of the mass attached to the spring, we can use the following formula:
x(t) = A * cos(ωt + φ),
where x(t) is the position of the mass at time t, A is the amplitude of the motion, ω is the angular frequency, t is the time, and φ is the phase constant.
Given information:
Mass (m) = 50 g = 0.05 kg,
Spring constant (k) = 0.10 N/m,
Initial position (x₀) = 14 cm to the right of the centre of oscillation O,
Initial velocity (v₀) = 23 cm/s away from O.
First, let's calculate the amplitude of the motion (A). Since the mass is attached to the spring, the amplitude is given by:
A = x₀ = 14 cm = 0.14 m.
Next, we need to find the angular frequency (ω). For a mass-spring system, the angular frequency is given by:
ω = √(k/m).
Substituting the values, we get:
ω = √(0.10 N/m / 0.05 kg) = √2 rad/s.
Now, let's find the phase constant (φ) using the initial velocity (v₀). Since the motion is initially away from O, the phase constant can be determined as:
φ = arccos(v₀ / (ωA)).
Substituting the values, we have:
φ = arccos(23 cm/s / (√2 rad/s * 0.14 m)).
Simplifying the equation, we get:
φ ≈ arccos(23 / (√2 * 0.14)) ≈ 1.446 rad.
Finally, we can write the position-time equation for the SHM:
x(t) = 0.14 * cos(√2t + 1.446).
Therefore, the position-time equation for the given SHM is x(t) = 0.14 * cos(√2t + 1.446).
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Starting from rest, a car accelerates 10 seconds with acceleration 4 m/s2 along straight line path . Determine its final velocity and the distance travelled during this time.
Its final velocity of the car is 40 m/s and the distance travelled during 10-second period is 200 meters.
To determine the final velocity and distance traveled by the car, we can use the following kinematic equations:
Final velocity (v):
v = u + at
Distance traveled (s):
s = ut + (1/2)at^2
Where:
u is the initial velocity (assumed to be 0 m/s since the car starts from rest).
a is the acceleration (4 m/s² in this case).
t is the time period (10 seconds in this case).
Calculating the final velocity (v):
v = u + at
v = 0 + 4 * 10
v = 40 m/s
Therefore, the final velocity of the car is 40 m/s.
Calculating the distance traveled (s):
s = ut + (1/2)at^2
s = 0 * 10 + (1/2) * 4 * (10)^2
s = 0 + 0.5 * 4 * 100
s = 0 + 200
s = 200 m
Therefore, the distance traveled by the car during the 10-second period is 200 meters.
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How much energy (in joules) is required to bring a particle having charge 49 micro-coulomb from very far away to a point that is at a distance R=2 meter from another (stationary) charge of 31 micro-coulomb.
The energy required to bring the particle from infinity to a distance of 2 meters from the stationary expense is 6.82 x 10^-3 Joules.
The electric potential energy between two charges is given by the formula:
U = k * q₁ * q₂ / r
where
k is Coulomb's constant,
q₁ and q₂ are the magnitudes of the charges,
r is the distance between them
The electric potential energy required to bring a charge from infinity to a distance R from another charge is given by the formula:
U = k * q₁ * q₂ / R
The total amount of energy required to bring a particle having charge 49 µC from infinity to a distance of 2 meters from another (stationary) charge of 31 µC can be calculated as follows:
U = k * q₁ * q₂ / R
where
k = 9 x 10^9 Nm²/C² is Coulomb's constant
q₁ = 49 µC is the charge of the particle
q₂ = 31 µC is the charge of the stationary charge
R = 2 m is the distance between the two charges
Therefore,
U = (9 x 10^9 Nm²/C²) x (49 x 10^-6 C) x (31 x 10^-6 C) / 2 m= 6.82 x 10^-3 J
Thus, the energy required to bring the particle from infinity to a distance of 2 meters from the stationary charge is 6.82 x 10^-3 Joules.
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While standing on top of a bridge over water, you drop a rock from rest, and it hits the water 1.5 seconds later. Then, you throw another rock straight down with an initial speed of 5.0 m/s. How long does the second rock take to hit the water? [Use the value g=9.8 m/s
2
for the acceleration of gravity.]
It takes about 0.51 seconds for the second rock to hit the water.
Acceleration due to gravity, g = 9.8 m/s^2
Time taken by the first rock to hit the water, t1 = 1.5 s
Initial speed of the second rock, u = 5.0 m/s
We need to calculate the time taken by the second rock to hit the water, t2.
To solve the above problem, we can use the formula:
v = u + gt
where
v = final velocity
u = initial velocity
t = time taken
g = acceleration of gravity
Since the rock is thrown straight down, the acceleration due to gravity acts downwards. Therefore, taking downwards as the positive direction, we can take the acceleration due to gravity as positive and write the above equation as:
v = u + gt
=> 0 = 5.0 + gt
t = -5.0 / g = -5.0 / 9.8≈ -0.510 s
Thus, it takes about 0.51 seconds for the second rock to hit the water.
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It was shown in Example 21.11 (Section 21.5 ) in the textbook that the electric field due to an infinite line of charge is perpendicular to the line and has magnitude E=λ/2πϵ
0
r. Consider an imaginary cylinder with a radius of r=0.160 m and a length of l=0.415 m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is λ=6.85μC/m. - Part B What is the flux through the cylinder if its radius is increased to r=0.535 m ? Express your answer in newton times meters squared per coulomb. - Part C What is the flux through the cylinder if its length is increased to l=0.865 m ? Express your answer in newton times meters squared per coulomb.
(a) The formula for the electric field due to an infinite line of charge is given as E = λ / (2πε₀r), where λ is the charge per unit length, ε₀ is the permittivity of free space, and r is the distance from the line of charge.
(b)The Electric flux through the cylinder when its radius is increased to r = 0.535 m is approximately 3.22 N·m²/C.
(c)The flux through the cylinder when its length is increased to l = 0.865 m is approximately 6.69 N·m²/C.
b) The surface area of the cylinder is given by A = 2πrh, where r is the radius and h is the height (length) of the cylinder.
Substituting the values, A = 2π(0.535)(0.415) = 1.11 m².
The flux through the cylinder is then given by Φ = EA, where E is the electric field and A is the surface area.
Φ = (λ / (2πε₀r)) * (2πrh) = λh / ε₀.
Substituting the given values, λ = 6.85 μC/m and ε₀ = 8.85 x 10^-12 C²/(N·m²), and h = 0.415 m, we can calculate the flux:
Φ = (6.85 x 10^-6 C/m)(0.415 m) / (8.85 x 10^-12 C²/(N·m²))
≈ 3.22 N·m²/C.
(c)The flux through the cylinder when its length is increased to l = 0.865 m is approximately 6.69 N·m²/C.
Φ = (6.85 x 10^-6 C/m)(0.865 m) / (8.85 x 10^-12 C²/(N·m²))
≈ 6.69 N·m²/C.
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Like 7−4 A three-phase 60−Hz induction motor runs at 710rpm at no load and at 670rpm at full load. a) How many poles does this motor have? b) What is the slip at rated load? c) What is the speed at one-quarter of the rated shaft torque? (You may assume the torque-speed curve is linear in this region). d) What is the rotor's electrical frequency at one-quarter of the rated shaft torque? 2. a) 10 b) 6.94.% c) 700⋅rpm d) 1.37⋅Hz
a) To determine the number of poles, we can use the formula:
Poles = (120 * Frequency) / RPM
Given that the frequency is 60 Hz and the no-load speed is 710 rpm, we can substitute these values into the formula:
Poles = (120 * 60) / 710
Simplifying this equation, we find that the motor has 10 poles.
b) The slip at rated load can be calculated using the formula:
Slip = (N_sync - N_full_load) / N_sync
Where N_sync is the synchronous speed and N_full_load is the full load speed. The synchronous speed can be calculated as:
N_sync = (120 * Frequency) / Poles
Substituting the given values into the equation, we have:
N_sync = (120 * 60) / 10
N_sync = 720 rpm
Now we can calculate the slip:
Slip = (720 - 670) / 720
Slip = 0.0694 or 6.94%
c) To find the speed at one-quarter of the rated shaft torque, we can assume that the torque-speed curve is linear in this region. So, if the full load speed is 670 rpm, the speed at one-quarter of the rated shaft torque would be:
Speed = Full Load Speed - (3/4 * (Full Load Speed - No Load Speed))
Speed = 670 - (3/4 * (670 - 710))
Speed = 670 - (3/4 * (-40))
Speed = 670 + 30
Speed = 700 rpm
d) The rotor's electrical frequency at one-quarter of the rated shaft torque can be calculated using the formula:
Frequency = Slip * Frequency
Substituting the given values, we have:
Frequency = 0.0694 * 60
Frequency = 4.16 Hz
Therefore, the correct answers are:
a) 10 poles
b) 6.94% slip
c) 700 rpm
d) 4.16 Hz
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A small marble is rolling down on an incline. The marble was released from rest when the timer was started. The distance travelied by the marale as the Aunction of time is shown in the figure. What is
To determine the acceleration of the marble rolling down an incline, we need to analyze the change in distance traveled with respect to time. By calculating the change in velocity over equal time intervals, we can determine the acceleration of the marble. The acceleration of the marble is approximately 3.428 m/s^2.
To find the acceleration, we can use the equation \(a = \frac{{\Delta v}}{{\Delta t}}\), where \(a\) represents acceleration, \(\Delta v\) represents the change in velocity, and \(\Delta t\) represents the change in time.
By observing the given figure, we can determine the change in distance traveled over equal time intervals. For example, we can calculate the change in distance between points 1 and 2, points 2 and 3, and so on. Dividing each change in distance by the time interval (which is constant), we obtain the average velocity for each interval.
To estimate the instantaneous velocity at each point, we assume that the average velocity over each interval applies to the midpoint of that interval. We can then calculate the change in velocity between adjacent points, such as the change in velocity between points 1 and 2, points 2 and 3, and so on.
Finally, by dividing the change in velocity by the constant time interval, we obtain the acceleration. Based on the given data, the calculated acceleration is approximately 3.428 m/s^2.
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The complete question is:
A small marble is rolling down on an incline. The marble was released from rest when the timer was started. The distance travelled by the marble as the function of time is shown in the figure. dicml 0 1 2 3 4 5 6 7 8 9 10 What is the acceleration of the marble? Please, note that the curve passes through at least one grid intersection point. 3.428 m/s^2 Submit Answer You have entered that answer before Incorrect. Tries 3/99 Previous Tries
A small rolling marble's speed on an incline depends on several factors such as the influence of friction and gravitational forces. By considering these, it's possible to calculate the marble's acceleration and potential velocity at any given time.
Explanation:The question pertains to the dynamics of a rolling marble on an incline. By observing the given scenarios and relevant physics concepts, we can determine a few key factors relating to the marble's movement. First, the marble's speed can be calculated using the average speed equation, which requires values for distance travelled and time interval. Secondly, we can infer from a given example scenario that a marble, or any object for that matter, will slide down a frictionless incline with the same acceleration, regardless of its mass, as long as the angle is the same.
Using these pieces of information, when examining the movement of a marble on an incline, you should always first consider the impact of friction and gravitational forces. This will allow you to calculate the exact acceleration and potential velocity of the marble at any given time.
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A particle is moving in a straight line along the x-axis with a constant acceleration. The equation of motion is
x(t)=(2.0m)−(4.0ms)t+(1.0ms2)t2
At time t = 0 s, the particle is:
a)moving to the right with a constant speed.
b)moving to the right with an increasing speed.
c)moving to the right with a decreasing speed.
d)moving to the left with a constant speed.
e)moving to the left with an increasing speed.
f)moving to the left with a decreasing speed.
g)at rest
The particle is moving to the right with a decreasing speed. Option (c) is correct.
The equation of motion given is as follows:
x(t)=(2.0m)−(4.0ms)t+(1.0ms²)t²
At time t = 0 s, x(0) = 2.0 m.
Therefore, the particle is initially at a distance of 2.0 m to the left of the origin (x=0).
The particle's velocity can be found by differentiating the equation of motion with respect to time:t= 0 = -4 m/sFrom this, we know that the particle is moving to the right.
The acceleration of the particle is the derivative of the velocity with respect to time:
a= -4 m/s²
The acceleration is negative. This indicates that the particle is moving with decreasing speed. Therefore, option (c) is correct.
Explanation:
The speed of the particle can be determined by taking the absolute value of its velocity. Here, the velocity is negative, so the speed is positive. However, the speed is decreasing, which is determined by the negative acceleration. Thus, the particle is moving to the right with a decreasing speed. Option (c) is correct.
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SERCP11 16.1.P.007.MI. Oppositely charged parallel plates are separated by 5.09 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? N/C (b) What is the magnitude of the force on an electron between the plates? N (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.12 mm from the positive plate?
Given information:Potential difference between plates = 600 VDistance between the plates = 5.09 mm = 5.09 × 10⁻³ mDistance of electron from the positive plate = 2.12 mm = 2.12 × 10⁻³ m(a) Magnitude of the electric field between the plates.
The electric field between two oppositely charged plates is uniform and is given by:E = V/dwhere,E = electric fieldV = potential difference between the platesd = distance between the platesPutting the given values, we get:E
= 600 V/5.09 × 10⁻³ mE = 1.18 × 10⁵ N/C
Thus, the magnitude of the electric field between the plates is 1.18 × 10⁵ N/C.(b) Magnitude of the force on an electron between the plates.
The force acting on an electron due to electric field E is given by,F = E × ewhere,F = force on the electronE = electric fielde = charge on an electron = -1.6 × 10⁻¹⁹ C (as electron is negatively charged)Putting the values, we get:
F = 1.18 × 10⁵ N/C × (-1.6 × 10⁻¹⁹ C)F = -1.88 × 10⁻¹⁴ N
(as the force is negative, it means the force is in the opposite direction to the direction of the electric field).
Thus, the magnitude of the force on an electron between the plates is 1.88 × 10⁻¹⁴ N.(c) Work done on the electron to move it to the negative plate:When a force is applied on an object, it moves in the direction of the force. The work done is given by:W = F × dwhere,W = work doneF = force on the electrond = distance moved by the electronTo move the electron from the positive plate to the negative plate, the electron has to move a distance of
5.09 - 2.12 = 2.97 mm = 2.97 × 10⁻³ mPutting the values, we get:
W = -1.88 × 10⁻¹⁴ N × 2.97 × 10⁻³ mW = -5.59 × 10⁻¹⁷ J
Thus, the work done on the electron to move it to the negative plate is -5.59 × 10⁻¹⁷ J.
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In an aluminum pot, .719 kg of water at 100 C boils away in four minutes. The bottom of the pot is 1.64 x 10 ^-3 m tick and has a surface area of .0168 m^2. to prevent water from boiling too rapidly, a stainless steel plate has been placed between the pot and the heating element. the plate is 1.29 x 10^-3 m thick, and its are matched that of the pot. Assuming that heat is conducted into the water only through the bottom of the pot, find the temperature in degrees C at the steel surface in contact with the heating element.
Correct answer: 209.4 °C.
Given data:
Mass of water = 0.719 kg
Initial temperature of water = 100°C
Final temperature of water = 100°C
Density of aluminum = 2700 kg/m³
Density of stainless steel = 7900 kg/m³
Specific heat of water = 4190 J/(kgK)
Thickness of aluminum pot = 1.64 × 10⁻³ m
Surface area of aluminum pot = 0.0168 m²
Thickness of stainless steel plate = 1.29 × 10⁻³ m
Let's find out the heat absorbed by the water to boil it. We can use the following formula for this:
[tex]\[Q = m \times c \times \Delta T\][/tex]
where,
Q = heat absorbed by the water
m = mass of water
c = specific heat of water
ΔT = change in temperature
∴ Q = 0.719 × 4190 × (100 - 100) J
Q = 0 J (as there is no change in temperature)
As there is no change in temperature, the heat absorbed by the water is zero. Therefore, the heat absorbed by the pot and plate is equal to the heat released by the heating element. We can use the following formula for this:
[tex]\[Q = kA\Delta T/ d\][/tex]
where,
Q = heat absorbed or released
k = thermal conductivity
A = surface area
d = thickness of the material
ΔT = temperature difference between two sides
Let's calculate the heat absorbed by the pot. We can use the following formula for this:
[tex]\[Q = kA\Delta T/ d\][/tex]
where,
Q = heat absorbed by the aluminum pot
k = thermal conductivity of aluminum
A = surface area of the aluminum pot
d = thickness of the aluminum pot
ΔT = temperature difference between two sides
∴ Q = 237 × 0.0168 × (T - 100)/(1.64 × 10⁻³) J
Let's calculate the heat absorbed by the steel plate. We can use the following formula for this:
[tex]\[Q = kA\Delta T/ d\][/tex]
where,
Q = heat absorbed by the steel plate
k = thermal conductivity of steel
A = surface area of the steel plate
d = thickness of the steel plate
ΔT = temperature difference between two sides
∴ Q = 16.2 × 0.0168 × (T - 100)/(1.29 × 10⁻³) J
As the heat absorbed by the pot is equal to the heat absorbed by the steel plate, we can equate the above two equations:
[tex]\[237 × 0.0168 × (T - 100)/(1.64 × 10⁻³) = 16.2 × 0.0168 × (T - 100)/(1.29 × 10⁻³)\][/tex]
∴ 120,123(T - 100)/(1.64 × 10⁻³) = 210,233(T - 100)/(1.29 × 10⁻³)
∴ 2.172(T - 100) = 3(T - 100)
∴ 2.172T - 217.2 = 3T - 300
∴ T = 209.4°C
Correct answer: 209.4 °C.
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What is the torque (in Nm) on a 0.4-kg basketball with a 0.3 m diameter as it rolls down a hill without slipping at an acceleration of 5 m/s2?
The torque on a 0.4 kg, 0.3 m diameter basketball rolling down a hill at a speed of 5 m/s² without sliding is 0.5994 Nm.
To calculate the torque on the basketball, we need to use the equation: Torque = moment of inertia * angular acceleration
First, we need to find the moment of inertia of the basketball.
The moment of inertia of a solid sphere can be calculated using the formula:
I = (2/5) * m * r²
Where, m is the mass of the basketball and r is the radius of the basketball.
As per data, the basketball has a diameter of 0.3 m, the radius is half of that, so r = 0.15 m.
Plugging in the values, we get:
I = (2/5) * 0.4 kg * (0.15 m)²
I = 0.018 kg * m²
Next, we need to find the angular acceleration of the basketball.
The angular acceleration can be found using the formula:
angular acceleration = linear acceleration / radius
As per data, the linear acceleration is 5 m/s² and the radius is 0.15 m, we have:
angular acceleration = 5 m/s² / 0.15 m
= 33.33 rad/s²
Now, we can calculate the torque using the equation:
Torque = 0.018 kg * m² * 33.33 rad/s²
= 0.5994 Nm
Therefore, the torque is 0.5994 Nm.
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A particular spring, with an equilibrium length of 40 cm, is
stretched to a length of 70 cm
as a force of 210 N is applied. What is the spring constant?
The spring constant of the particular spring can be determined using Hooke's Law, which states that the force applied to a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, Hooke's Law can be represented as F = k * x, where F is the force applied, k is the spring constant, and x is the displacement.
Given:
Equilibrium length (x₀) = 40 cm
Stretched length (x) = 70 cm
Force applied (F) = 210 N
To find the spring constant (k), we need to calculate the displacement (x - x₀). Therefore:
Displacement (x - x₀) = 70 cm - 40 cm = 30 cm = 0.3 m
Using Hooke's Law, we can rearrange the formula to solve for the spring constant (k):
k = F / (x - x₀)
k = 210 N / 0.3 m
Calculating the spring constant:
k = 700 N/m
Therefore, the spring constant of the particular spring is 700 N/m.
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An infinite line charge of uniform linear charge density = -3.7 C/m lies parallel to the y axis at x = 0 m. A point charge of 1.7 C is located at x = 1.0 m, y = 2.0 m.
Find the x component of the electric field at x = 2.0 m, y = 1.5 m.
________________ kN/C
The x-component of the electric field at the given point is approximately 3.78 × 10^6 N/C, and the y-component of the electric field at the given point is approximately 3.78 × 10^6 N/C.
To find the x- and y-components of the electric field at the given point, we can use the principle of superposition. We need to consider the electric field contributions from both the infinite line charge and the point charge.
The electric field due to an infinite line charge at a perpendicular distance r from the line charge is given by:
E_line = (λ / (2πε₀r)) * (cosθ₁ - cosθ₂)
where:
λ is the linear charge density of the line charge,
ε₀ is the permittivity of free space,
r is the distance from the line charge,
θ₁ is the angle between the line charge and the line connecting the point to the line charge,
θ₂ is the angle between the line charge and the line connecting the point to the line charge.
In this case, the linear charge density (λ) is -1.3 μC/m and the distance from the line charge (r) is √((1.5-(-2))^2 + (2.0-1.5)^2) = √(3.5^2 + 0.5^2) = √12.5 ≈ 3.54 m. The angles θ₁ and θ₂ can be calculated using the geometry of the problem.
Next, we need to calculate the electric field due to the point charge at the given point. The electric field due to a point charge is given by:
E_point = (k * q) / r²
where:
k is the electrostatic constant (9 x 10^9 N m²/C²),
q is the charge of the point charge,
r is the distance from the point charge.
In this case, the charge of the point charge (q) is 2.1 μC and the distance from the point charge (r) is √((2.0-1.5)^2 + (2.0-1.5)^2) = √(0.5^2 + 0.5^2) = √0.5 ≈ 0.71 m.
Now we can calculate the x- and y-components of the electric field at the given point by summing up the contributions from the line charge and the point charge:
E_x = E_line_x + E_point_x
E_y = E_line_y + E_point_y
where E_line_x and E_line_y are the x- and y-components of the electric field due to the line charge, and E_point_x and E_point_y are the x- and y-components of the electric field due to the point charge.
Finally, we can calculate the x- and y-components of the electric field at the given point.
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How does a protoplanet differ from a moon? Choose one: A. Protoplanets orbit the Sun: moons orbit planets. B. Protoplanets orbit Jovian planetsi moons orbit both terrestrial and Jovian planets. C. Pro
Option A is correct. Protoplanets differ from moons in that they orbit the Sun, while moons orbit planets or both terrestrial and Jovian planets.
Protoplanets are celestial bodies that are in the early stages of formation as they accumulate matter in a disk around a star, such as the Sun. They are the building blocks of planets and are characterized by their larger size compared to moons. Protoplanets primarily orbit the Sun, similar to how planets orbit the Sun.
On the other hand, moons are natural satellites that orbit around planets or both terrestrial (rocky) and Jovian (gas giant) planets. Moons can vary in size and are often formed from the remnants of planetary formation or through capture by a planet's gravitational pull. Therefore, the key distinction lies in the orbits: protoplanets orbit the Sun, while moons orbit planets or both terrestrial and Jovian planets.
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Which of the following characteristics will create the lowest speed of sound? Low stiffness, high density High stiffness, high density Low stiffness, high impedance Low stiffness, low density Low stiffness, low impedance Question 11 What causes the near field of image to be brighter than the far field? Stronger sound wave in the far field Less refraction sound in the of far field More attenuation of sound in the far field More refraction sound in the of far field Propagation speed of sound
The characteristics that create the lowest speed of sound are low stiffness and high density.
Sound is defined as mechanical waves that require a medium to travel from one point to another.
The speed of sound is the speed at which sound waves propagate through any medium;
it depends on the characteristics of that medium.
The speed of sound is inversely proportional to the density and proportional to the stiffness of the medium.
Low stiffness and high density of the medium will produce the lowest speed of sound since both properties are inversely proportional to the speed of sound.
A low stiffness implies that the medium is soft and flexible, and this reduces the speed of sound because sound waves travel much slower in soft materials.
High density implies that the particles in the medium are tightly packed, thus reducing the speed of sound since particles must collide to transmit the sound wave.
Hence, low stiffness and high density produce the lowest speed of sound.
The near field of an image is always brighter than the far field because more refraction of sound occurs in the near field than the far field.
The sound wave undergoes refraction as it travels through a medium with varying temperatures, densities, and pressures.
In the near field, the sound waves experience more refraction, making the waves to bend more, thus increasing the amplitude of the wavefronts and making the field brighter.
This is different from the far field, where the sound wave undergoes less refraction, leading to a lower amplitude, making the field less bright.
Hence, the brightness of the near field is due to the amount of refraction sound waves experience, while the brightness of the far field is due to the propagation speed of sound.
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A He-Ne laser emits light with a wavelength of 610 nm. This is sent through a double slit with a slit spacing of 1.9 mm. The screen where the interference pattern appears is 5.5 m from the slot. Calculate the distance between the central maximum and the first bright spot. Please answer in word prg
The distance between the central maximum and the first bright spot in the interference pattern is approximately 1.14 meters. We can use the formula: dsinθ = mλ.
To calculate the distance between the central maximum and the first bright spot in an interference pattern produced by a double slit, we can use the formula:
dsinθ = mλ
where:
d is the slit spacing,
θ is the angle between the central maximum and the bright spot,
m is the order of the bright spot,
λ is the wavelength of the light.
Given:
Wavelength of light (λ) = 610 nm = 610 × 10^(-9) m
Slit spacing (d) = 1.9 mm = 1.9 × 10^(-3) m
Distance to the screen (L) = 5.5 m
We want to find the distance between the central maximum and the first bright spot, which corresponds to m = 1.
Rearranging the formula, we have:
θ = arcsin(m*λ/d)
θ = arcsin(1 * 610 × 10^(-9) m / 1.9 × 10^(-3) m)
Using a calculator, θ ≈ 0.197 rad
To find the distance between the central maximum and the first bright spot (x), we can use trigonometry:
x = L * tan(θ)
x = 5.5 m * tan(0.197 rad)
Using a calculator, x ≈ 1.14 m
Therefore, the distance between the central maximum and the first bright spot in the interference pattern is approximately 1.14 meters.
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Vector Addition 4: Find the magnitude and direction of the resultant of the following three force vectors: F
1
=(0.100 kg)×g=0.981Nat30
∘
F
2
=(0.200 kg)×g=1.96 N at 90
∘
F
3
=(0.300 kg)×g=2.94 N at 225
∘
using the following three procedures: a) Graphical. Use the polygon method. b) Analytical. Use the component method. c) Experimental. Use the force table.
a) Using polygon method, the magnitude of the resultant force is 4.11 N and the direction of the resultant is 12° (approx.) from the direction of the vector F1.
b) Using the component method, the magnitude of the resultant force is 1.28 N (approx.) and the direction of the resultant vector is -17.7° (approx.).
c) Using the force table, the magnitude of the resultant force is 1.33 N and the direction of the resultant vector is -16° (approx.) from the x-axis.
a) Graphical
Using polygon method
F1 = 0.100 kg × g = 0.981 N at 30°
F2 = 0.200 kg × g = 1.96 N at 90°
F3 = 0.300 kg × g = 2.94 N at 225°
For vector addition using the polygon method, we will follow the below procedure:
Step 1: Draw the first vector F1 to the scale.
Step 2: Draw the second vector F2 to the same scale, starting from the end of the vector F1.
Step 3: Draw the third vector F3 to the same scale, starting from the end of the vector F2.
Step 4: The resultant of the three vectors will be a line drawn from the start of the vector F1 to the end of the vector F3.
Step 5: Measure the length of the resultant R by using the scale. This will give us the magnitude of the resultant.
Step 6: Measure the angle between the resultant R and the first vector F1 using a protractor. This will give us the direction of the resultant.
The above steps are followed in the figure given below.
Now, we can see that the length of the line OR is 4.11 cm (measured using the scale).
We can calculate the magnitude of the resultant force by using the following formula:
R = (4.11 cm) × (1 N/cm)
R = 4.11 N
Thus, the magnitude of the resultant is 4.11 N.
Now, to find the direction of the resultant, we need to measure the angle θ between the resultant and the first vector F1 using a protractor. We get:
θ = 12° (approx.)
Thus, the direction of the resultant is 12° (approx.) from the direction of the vector F1.
b) Analytical.
Use the component method.
Let's calculate the horizontal and vertical components of all the three vectors:
F1 :Fx1 = F1 cos θ1 = (0.981 N) cos 30° = 0.850 N
Fy1 = F1 sin θ1 = (0.981 N) sin 30° = 0.490 N
F2 :Fx2 = F2 cos θ2 = (1.96 N) cos 90° = 0 N
Fy2 = F2 sin θ2 = (1.96 N) sin 90° = 1.96 N
F3 :Fx3 = F3 cos θ3 = (2.94 N) cos 225° = -2.077 N
Fy3 = F3 sin θ3 = (2.94 N) sin 225° = -2.077 N
Now, let's calculate the resultant horizontal and vertical components:
Rx = Fx1 + Fx2 + Fx3
Rx = (0.850 N) + 0 + (-2.077 N)
Rx = -1.227 N
Ry = Fy1 + Fy2 + Fy3
Ry = (0.490 N) + (1.96 N) + (-2.077 N)
Ry = 0.373 N
Now, we can calculate the magnitude and direction of the resultant vector using the following formulae:
Magnitude of resultant vector R = √(Rx^2 + Ry^2)
Magnitude of resultant vector R = √((-1.227 N)^2 + (0.373 N)^2)
Magnitude of resultant vector R = 1.28 N (approx.)
Direction of the resultant vector θ = tan^-1(Ry/Rx)
Direction of the resultant vector θ = tan^-1(0.373 N/ -1.227 N)
Direction of the resultant vector θ = -17.7° (approx.)
Thus, the magnitude of the resultant vector is 1.28 N (approx.) and the direction of the resultant vector is -17.7° (approx.).
c) Experimental.
Use the force table.
To find the resultant using the force table, we need to follow the below procedure:
Step 1: Set up the force table with the three pulleys located at angles of 30°, 90° and 225° to each other.
Step 2: Attach weights to the strings at the pulleys such that the tension in the strings is equal to the magnitude of the corresponding force vectors.
Step 3: Adjust the directions of the forces until the ring (to which the strings are attached) is at the center of the force table.
Step 4: Measure the direction of the ring from the x-axis using a protractor. This will give us the direction of the resultant.
Step 5: Measure the force shown on the force table using a scale.
This will give us the magnitude of the resultant.
The above steps are followed in the figure given below.
Now, we can see that the ring is located at the center of the force table. We can measure the magnitude of the resultant by reading the value of the force shown on the force table. We get:
R = 1.33 N (approx.)
Thus, the magnitude of the resultant vector is 1.33 N (approx.).
To find the direction of the resultant vector, we need to measure the angle between the x-axis and the direction of the ring using a protractor. We get:
θ = -16° (approx.)
Thus, the direction of the resultant vector is -16° (approx.) from the x-axis.
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Four capacitors are connected as shown in the figure below. (Let C=14.0μF.) (a) Find the equivalent capacitance between points aand b. μF (b) Calculate the charge on each capacitor, taking ΔVab=14.0 V. 20.0μFcapacitor 6.00μF capacitor μC 3.00μFcapacitor capacitor C
A The equivalent capacitance between points a and b is C_{eq2} , (b) The charges on each capacitor are [tex]Q_{eq2}, Q_{C}, Q_{6 \mu F}, and Q_{20 \mu F}.[/tex]
find the equivalent capacitance between points a and b, we can analyze the circuit and apply the rules for combining capacitors in series and parallel.
(a) Capacitors C, 6.00μF, and the 20.0μF capacitor are in parallel, so their equivalent capacitance (C_eq1) can be calculated as:
1/C_eq1 = 1/20.0μF + 1/6.00μF
Solving this equation will give us the value of C_eq1.
Next, the capacitor C_eq1 and the 3.00μF capacitor are in series, so their equivalent capacitance (C_eq2) can be calculated as:
C_eq2 = C_eq1 + 3.00μF
The equivalent capacitance between points a and b is C_eq2.
(b) To calculate the charge on each capacitor, we can use the formula:
Q = C * ΔV
where Q is the charge, C is the capacitance, and ΔV is the potential difference across the capacitor.
Using the given value of ΔVab = 14.0 V, we can calculate the charge on each capacitor as follows:
For capacitor C_eq2: Q_eq2 = C_eq2 * ΔVab
For capacitor C: Q_C = C * ΔVab
For capacitor 6.00μF: Q_6μF = 6.00μF * ΔVab
For capacitor 20.0μF: Q_20μF = 20.0μF * ΔVab
Substituting the respective capacitance values, we can find the charge on each capacitor.
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An electric blanket is connected to a 120 -V outlet and consumes \( 350 \mathrm{~W} \) of power. What is the resistance of the heater wire in the blanket? Number Units
The resistance of the heater wire in the electric blanket is 41.1 ohms.
Given that an electric blanket is connected to a 120 -V outlet and consumes 350 W of power.
The resistance of the heater wire in the blanket can be calculated using the following formula:
Resistance (R) = Voltage (V) / Current (I)From Ohm's law, we can derive the formula for the resistance as:
R = V/I
For a given electrical power P in watts (W), voltage V in volts (V) and current I in amperes (A), this can also be written as:
P = V × I
Substituting the given values of voltage and power into the equation P = V × I, we can solve for current:I = P/V= 350 W / 120 V= 2.92 A
Finally, substituting the values of voltage and current in the formula for resistance R = V/I, we get;R = V/I= 120 V / 2.92 A= 41.1 Ω
Therefore, the resistance of the heater wire in the electric blanket is 41.1 ohms.
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P4. (a) (3 point5) Calculate the total cross-sectional area of the major arteries given that; The average blood velocity in the major arteries is 4 cm/s, the radius of the aorta is 1.0 cm, the blood velocity in the aorta is 30 cm/s.
the total cross-sectional area of the major arteries is 0.4187 cm².
Given that the average blood velocity in the major arteries is 4 cm/s, the radius of the aorta is 1.0 cm, the blood velocity in the aorta is 30 cm/s.
To find out the total cross-sectional area of the major arteries we can use the law of continuity which states that the mass flow rate of a fluid in a pipe is constant over time and the equation for mass flow rate is given by the product of density, cross-sectional area, and velocity of the fluid.
The law of continuity equation is as follows:mass flow rate = density × cross-sectional area × velocity of the fluidQ = AvWhere,Q = mass flow rate
A = cross-sectional areav = velocity of the fluidSince mass flow rate is constant throughout a pipe, we can equate the mass flow rate of blood in the aorta to the mass flow rate of blood in all of the major arteries:Qaorta = Qmajor arteriesThis gives the equation:
Aaorta × vaorta = Amajor arteries × vmajor arteries
Aaorta = Amajor arteries × (vmajor arteries/vaorta)
Putting in the given values, the total cross-sectional area of the major arteries is as follows
:Aaorta = [tex]pi x (1.0)^2[/tex] = piAaorta = 3.14The cross-sectional area of the major arteries = 3.14 × (4/30) = 0.4187 cm²
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The density of mercury is 1.36×10
4
kg/m
3
. What is the mass of a 2.16×10
−7
m
3
sample of mercury? A) 0.0343 kg B) 0.00294 kg C) 2.94 kg D) 5.44 kg E) 6.29 kg 2. At standard temperature and pressure, carbon dioxide has a density of 1.98 kg/m
3
. What volume does 851 grams of carbon dioxide occupy at standard temperature and pressure? A) 0.43 m
3
B) 0.86 m
3
C) 1.7 m
3
D) 2.3 m
3
E) 4.8 m 3. What mass of water (at 4.0
∘
C ) can be contained in a rectangular box whose dimensions are 1.00 cm by 5.00 cm by 1.00 cm ? The density of water at 4.0
∘
C is 1.000×10
3
kg/m
3
. A) 5.0 g B) 10.0 g C) 25.0 g D) 50.0 g E) 0.25 kg 4. The density of gold is 19300 kg/m
3
. What is the mass of a gold sphere whose diameter is 0.52 m ? A) 875 kg B) 158 kg C) 947 kg D) 1070 kg E) 1420 kg
The mass is:
1. The mass is 2.94 × 10-3 kg (option b).
2. The volume is 0.43 m3 (option a).
3. The mass of water is 5 × 10-3 kg (option a).
4. The mass is 227.74 kg (option D).
1. The given density of mercury is 1.36×104 kg/m3. The mass of a 2.16×10−7 m3 sample of mercury can be calculated using the formula:
density = mass/volume
mass = density × volume
Therefore, mass = 1.36×104 kg/m3 × 2.16×10−7 m3= 2.94 × 10-3 kg (approximately 0.00294 kg) Therefore, option B is the correct answer.
2. The given density of carbon dioxide is 1.98 kg/m3. We need to find the volume that 851 grams of carbon dioxide occupies at standard temperature and pressure. Using the formula:
density = mass/volume
volume = mass/density
We have mass = 851 g = 0.851 kg
density = 1.98 kg/m3
Therefore, volume = 0.851 kg ÷ 1.98 kg/m3 = 0.43 m3
Therefore, option A is the correct answer.
3. The dimensions of the rectangular box are 1.00 cm by 5.00 cm by 1.00 cm. The volume of the rectangular box is given by:
volume = length × width × height= 1.00 cm × 5.00 cm × 1.00 cm= 5.00 cm3= 5.00 × 10-6 m3
Using the formula:
density = mass/volume
mass = density × volume
We have the density of water = 1000 kg/m3 and the volume of the rectangular box = 5.00 × 10-6 m3
Therefore, mass of water = 1000 kg/m3 × 5.00 × 10-6 m3= 5 × 10-3 kg
Therefore, the correct answer is option A, 5.0 g.
4. The diameter of the gold sphere is given to be 0.52 m. The radius can be found by dividing the diameter by 2. Therefore, radius, r = 0.52 m ÷ 2 = 0.26 m.
Using the formula of the volume of the sphere:v = (4/3) × π × r3
Therefore, v = (4/3) × 3.14 × (0.26 m)3= 0.0118 m3
Using the formula:density = mass/volume
mass = density × volume
We have the density of gold to be 19300 kg/m3
Therefore, mass = 19300 kg/m3 × 0.0118 m3= 227.74 kg (approx. 228 kg)
Therefore, the correct answer is option D, 1070 kg.
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The highest barrier that a projectile can clear is 11.8 m, when the projectile is launched at an angle of 23.0
∘
above the horizontal. What is the projectile's launch speed? Number Units
The projectile's launch speed is approximately 14.9 m/s when the highest barrier it can clear is 11.8 m at an angle of 23.0° above the horizontal.
Given, Barrier that a projectile can clear, h = 11.8 m, Angle made by the projectile with horizontal, θ = 23.0°, Kinematic equation used to find the launch speed of the projectile when it is launched at an angle above the horizontal is:
v² = u² + 2gh where u = initial velocity of the projectile, v = final velocity of the projectile after attaining a height h above the ground, g = acceleration due to gravity, h = height attained by the projectile above the ground
According to the problem, the projectile attains a maximum height of 11.8 m above the ground.
Therefore, using the above kinematic equation, we can write,0 = u² + 2gh => u² = -2gh
Putting the values of h and θ in the above equation, we get
u = √(2gh)
= √(2 * 9.8 * 11.8 * sin²23.0)
≈ 14.9 m/s
Therefore, the launch speed of the projectile is approximately 14.9 m/s.
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