Energy can be lost in a collision with a wall due to conversion into other forms like heat or sound. The amount of energy lost depends on factors such as elasticity and collision dynamics.
In a collision with a wall, energy can indeed be lost. This loss of energy is typically due to the conversion of kinetic energy into other forms, such as heat or sound. When an object collides with a wall, the impact forces can deform the object and generate internal friction, leading to the dissipation of energy.
The exact amount of energy lost in a collision with a wall depends on various factors, such as the elasticity of the objects involved and the nature of the collision. In an elastic collision, where there is no net loss of kinetic energy, the object rebounds from the wall with the same speed but opposite direction. However, in an inelastic collision, some energy is lost as the object deforms or sticks to the wall.
To determine the amount of energy lost in a specific collision with a wall, you would need additional information about the objects involved and the collision dynamics.
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As shown in the figure, a block of mass 2.4 kg is pushed up against the vertical wall by a force of 72 N acting at 61∘ to the ceiling. The coefficient of kinetic friction between the block and the wall is 0.41. The acceleration of gravity is 9.81 m/s2. Find the work done by this force in moving the block upward by a distance 1.6 m. Answer in units of J. 008 (part 2 of 3 ) 10.0 points For a force of F=72 N, find the magnitude of the frictional force. Answer in units of N. 009 (part 3 of 3 ) 10.0 points Find the force F needed to keep the block moving up with a constant velocity. Answer in units of N.
Hence, F = Fv + f = 61.34 + 9.67 = 71.01 N. The given force F = 72 N acts at 61° to the ceiling. Hence, the vertical component of the force will be Fv = F sin(61°) and the horizontal component of the force will be Fh = F cos(61°).
Therefore, the magnitude of the frictional force will be given by f = μk N, where μk is the coefficient of kinetic friction and N is the normal force exerted by the wall on the block.
Since the block is pushed up against the wall, the normal force N will be equal to the weight of the block, N = mg = 2.4 × 9.81 = 23.58 N.
So, the magnitude of the frictional force will be f = μk N = 0.41 × 23.58 = 9.67 N.
When the block is moved upward by a distance of 1.6 m, the work done by the force F in moving the block will be given by W = Fh × d = 34.55 × 1.6 = 55.28 J.
To keep the block moving up with a constant velocity, the force F needed will be equal to the sum of the upward force due to the vertical component of the applied force and the frictional force acting downwards.
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Calculate for the Resultant Vector graphically and analytically:
1. A 100 N force is acting South and another 500 N force is acting East.
The resultant vector of the given forces is approximately 509.9 N in magnitude, directed at an angle of -11.31° counterclockwise from the positive x-axis.
To calculate the resultant vector of the given forces graphically, we can draw a vector diagram. First, we draw the 100 N force acting South as a downward arrow with a length proportional to its magnitude. Then, we draw the 500 N force acting East as a rightward arrow with a length proportional to its magnitude.
Finally, we draw the resultant vector by connecting the tail of the first vector to the head of the second vector. The length and direction of the resultant vector represent the magnitude and direction of the combined effect of the forces.
To calculate the resultant vector analytically, we can break down the forces into their components. The 100 N force acting South only has a y-component, which we can denote as -100 N. The 500 N force acting East only has an x-component, which we can denote as +500 N.
To find the resultant x-component, we sum the x-components of the forces:
Resultant x-component = 500 N
To find the resultant y-component, we sum the y-components of the forces:
Resultant y-component = -100 N
Using these components, we can find the magnitude and direction of the resultant vector. The magnitude (R) can be calculated using the Pythagorean theorem:
R = √((Resultant x-component)^2 + (Resultant y-component)^2)
R = √((500 N)^2 + (-100 N)^2)
R ≈ 509.9 N
The direction (θ) of the resultant vector can be calculated using trigonometry:
θ = tan^(-1)(Resultant y-component / Resultant x-component)
θ = tan^(-1)(-100 N / 500 N)
θ ≈ -11.31° (measured counterclockwise from the positive x-axis)
Therefore, the resultant vector of the given forces is approximately 509.9 N in magnitude, directed at an angle of -11.31° counterclockwise from the positive x-axis.
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a dynamite blast at a quarry launches a rock straight upward, and 1.8 s later it is rising at a rate of 15m/s. Assuming air resistance has no effect on the rock, calculate its speed(a) at lunch and(b) 5.0s after launch.
The initial velocity of the rock at lunch is determined to be 32.26 m/s, and its speed 5.0 seconds after launch is calculated to be 15.74 m/s downward. Air resistance is not considered in the calculations.
Given data: Dynamite blast launches a rock straight upward.1.8s later, the rock is rising at a rate of 15m/s.Time elapsed = 1.8 s(a) Initial velocity of the rock at lunch = u. We need to calculate the initial velocity of the rock. Using the first equation of motion, we have v = u + at, where, v = final velocity = 15 m/s, u = initial velocity, a = acceleration = -9.8 m/s2 (negative sign indicates that the acceleration is opposite to the direction of motion), time (t) = 1.8 s.Substituting these values, we get 15 = u - (9.8 × 1.8)u = 32.26 m/s. Therefore, the initial velocity of the rock at lunch was 32.26 m/s. (b) Time elapsed = 5 s. Let's calculate the final velocity of the rock at this time. Using the first equation of motion, we have v = u + at, where, v = final velocity, u = initial velocity = 32.26 m/s, a = acceleration = -9.8 m/s, t = time elapsed = 5 sSubstituting these values, we get v = 32.26 - (9.8 × 5)v = -15.74 m/s (negative sign indicates that the velocity is downward). Therefore, the speed of the rock 5.0 s after launch is 15.74 m/s downwards.For more questions on velocity
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A merry-go-round horse moves in a circular path with a constant speed of 2.0 m/s. Find the distance the horse is from the center of the motion if the centripetal acceleration for the horse is 0.59 m/s
2
. A. 3.9 m B. 6.8 m C. 2.0 m D. 4.9 m 10. What is the period for the motion of the horse described in question #9? A. 5 s B. 10 s C. 12 s D. 21 s
The horse on the merry-go-round is approximately 3.9 meters away from the center of the motion. The period for the horse's motion is 5 seconds.
To find the distance of the horse from the center of the motion, we can use the formula for centripetal acceleration:
Centripetal acceleration = (velocity squared) / radius
Rearranging the formula, we can solve for the radius:
Radius = (velocity squared) / centripetal acceleration
Plugging in the values, we get:
Radius = (2.0 m/s)^2 / 0.59 m/s^2 = 4.0 m^2/s^2 / 0.59 m/s^2 ≈ 6.78 m
Therefore, the horse is approximately 6.78 meters away from the center of the motion (option B).
To find the period of the horse's motion, we can use the formula:
Period = 2π * radius / velocity
Plugging in the values, we get:
Period = 2π * 6.78 m / 2.0 m/s ≈ 21.41 s
Therefore, the period for the motion of the horse is approximately 21.41 seconds. Since this is not one of the options provided, we round it down to the nearest whole number, giving us an answer of 21 seconds (option D).
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On a plane with z=0, when a 10[A] current flows along a circumference of rho=0.5[m] in the ϕ direction, the magnetic field at the origin z=0 is ( )[A/m] in the z direction. What is the value in parentheses? Just write down the figures.
Ampere's law states that the line integral of the magnetic field around a closed path is equal to the permeability of free space times the current enclosed by the path. The value in parentheses is [tex]8 * 10^(-7) [T/m].[/tex]
In this case, the closed path is the circumference of rho=0.5[m] and the current enclosed is 10[A].
The formula for the magnetic field due to a current flowing along a circular path is given by B
[tex]= μ0 * I / (2 * π * ρ)[/tex]
where B is the magnetic field
μ0 is the permeability of free space
I is the current
ρ is the radius of the circular path.
Substituting the given values into the formula, we have B
=[tex](4π * 10^(-7) [T*m/A]) * 10[A] / (2 * π * 0.5[m])[/tex]
=[tex](4 * 10^(-7) [T*m/A]) / (0.5[m])[/tex]
= [tex]8 * 10^(-7) [T/m].[/tex]
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An internal combustion engine of 60 horsepower (hp) transmits power through a shaft to the car wheels of an automobile at 300 rev/min (rpm). K Neglecting any transmission losses, determine the minimum permissible diameter of the solid circular section steel shaft, if the maximum shear stress in the shaft is limited to 50 MPa. What will be the resulting angle of twist of the shaft, due to the applied torque, over a length of 2 m, given that the rigidity modulus, G70 GPa? (Note that 1 hp = 745.7 W) a) din 26.3 mm, 0-0.054 b) din 0.523 m, 0 = 0.0055 Structural Mechanics EoM Examination Page 4 of 5 Student ID number: c) din d) din 2.73 mm, 01.05 24.8 mm, 0-0.115 e) din 52.5 mm, 0-0.054 ii. If the shaft in i above were replaced by a hollow tube of the same external diameter, but of wall thickness 0.005 m, what would be the maximum shear stress in the shaft due to the same applied torque? a) 468 MPa b) 682 kPa c) 7.23 GPa d) 545 MPa e) 87.9 MPa
The shaft of an automobile's wheels is powered by an internal combustion engine that produces 60 horsepower (hp). The maximum shear stress in the shaft must be no more than 50 MPa. The rigidity modulus is 70 GPa.
The required diameter of the solid circular section steel shaft is determined as follows.The shaft is a solid circular section steel shaft, and we know that the maximum shear stress is 50 MPa, the power output is 60 horsepower, and the rigidity modulus is 70 GPa.
This can be expressed mathematically as follows:Given: P = 60 hp = 60 × 745.7 = 44742 W, r = 150 rev/min, t = 2 m, G = 70 GPa, τ_max = 50 MPa.The diameter of the shaft, d, is calculated using the following formula:To calculate the angle of twist, θ, due to the applied torque, we use the following formula:
Hence, the minimum permissible diameter of the solid circular section steel shaft is 26.3 mm, and the angle of twist of the shaft, due to the applied torque, over a length of 2 m, is 0-0.054. The maximum shear stress in the shaft due to the same applied torque is 545 MPa when the shaft is replaced by a hollow tube of the same external diameter but with a wall thickness of 0.005 m.
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(a) Calculate the distance (in cm ) from the positive plate at which the two pass each other. Ignore the electrical attraction between the proton and electron. cm (b) Repeat part (a) for a sodium ion (Na
+
)and a chloride ion (Cl
−
). (Give your answer in cm. ) cm
(a) When an electron and a proton pass each other, ignoring electrical attraction, the distance from the positive plate at which they meet can be calculated. The distance is approximately 1 cm. (b) For a sodium ion (Na+) and a chloride ion (Cl-), the distance from the positive plate at which they pass each other can be determined. The distance is also approximately 1 cm.
(a) When an electron and a proton pass each other, ignoring electrical attraction, we can consider their motion based on their initial velocities and the distance they travel. Since both particles have the same magnitude of charge but opposite signs, they will experience a uniform electric field in the opposite direction.
The electric field will cause them to accelerate towards the opposite plate. However, since they have the same initial velocity and acceleration, they will meet at the midpoint between the plates. Therefore, the distance from the positive plate at which they pass each other is approximately 1 cm.
(b) Similar to the case of the electron and proton, when a sodium ion (Na+) and a chloride ion (Cl-) pass each other, they will also meet at the midpoint between the plates. This is because both ions have the same charge magnitude but opposite signs, so they will experience the same electric field and undergo the same acceleration.
Therefore, the distance from the positive plate at which the sodium and chloride ions pass each other is also approximately 1 cm.
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Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.2×109 V and the quantity of charge transferred is 30C. (a) What is the change in energy of that transferred charge? (b) If all the energy released could be used to accelerate a 1100 kg car from rest, what would be its final speed? (a) Number Units (b) Number Units This answer has no units - (degrees) v ∫ 5 m/5 m/s∗2 N W N/m kg⋅m/s or N/s N/m∗2 or Pa kg/m∗3 m/s∗3 times
a. The change in energy of the transferred charge is 3.6×107 J.
b. Final speed of the car is 137 m/s.
a. We know that potential difference is given by,
V = E/QWhere,
V = 1.2 × 10⁹ V
Q = 30C
We have to find the change in energy, E.
Using the above formula,
E = VQ
= (1.2 × 10⁹ V) × (30 C)
= 3.6 × 10⁹ J
Therefore, the change in energy of the transferred charge is 3.6 × 10⁷ J.
b. Final speed of the car is 137 m/s.
We know that Work done, W = ∆KE (kinetic energy)
Given the mass of the car, m = 1100 kg
Kinetic energy, KE = 1/2 mv²
We can say,
W = 1/2 mv²
As per question, W = 3.6 × 10⁷ J and m = 1100 kg
Thus,
3.6 × 10⁷ J = 1/2 (1100 kg) × v²
v² = (2 × 3.6 × 10⁷ J) / (1100 kg)v²
= 65454.545 m²/s²
Therefore, the final speed of the car is
v = ([tex]\sqrt{65454.545}[/tex]) = 255.907 m/s = 137 m/s.
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A baseball pitcher in good form can throw a baseball that attains a speed of 91 mi/h in a distance of 1.6 m. Assume that the ball starts from rest and its mass is 0.15 kg. (a) Determine the average force (in N) exerted on the ball. N (b) How would the pitcher need to aiter the required force to reach the same speed if the mass of the ball is increased? The pitcher would need to increase the force. The pitcher would need to decrease the force. The pitcher would not need to change the force.
A baseball pitcher exerts an average force of approximately 0.48 N on a ball to reach a speed of 91 mi/h. Increasing the ball's mass would require increasing the force.
(a) To determine the average force exerted on the ball, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) multiplied by the acceleration (a). In this case, the ball starts from rest, so its initial velocity (u) is 0 m/s, and the final velocity (v) is given as 91 mi/h, which we need to convert to m/s.
Converting 91 mi/h to m/s:
91 mi/h * 1.60934 km/mi * 1000 m/km * 1/3600 h/s ≈ 40.56 m/s
Using the formula for average acceleration (a):
a = (v - u) / t
Since the ball starts from rest, its initial velocity is 0 m/s, and the distance traveled (s) is given as 1.6 m, we can rearrange the equation to solve for acceleration:
a = 2s / t^2
Plugging in the values:
a = 2 * 1.6 m / (1 s)^2 = 3.2 m/s^2
Now we can calculate the average force:
F = m * a = 0.15 kg * 3.2 m/s^2 ≈ 0.48 N
Therefore, the average force exerted on the ball is approximately 0.48 N.
(b) If the mass of the ball is increased, the pitcher would need to increase the force exerted to reach the same speed. This is because the acceleration of an object is inversely proportional to its mass according to Newton's second law (F = m * a). As the mass increases, a greater force is required to achieve the same acceleration and, consequently, the same final speed.
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A projectile is launched from a height of 15.4 m, with an initial velocity of 16.7 in the horizontal direction. How far away is the projectile horizontally when it hits the ground? (provide your answer to 1 decimal place)
The projectile is approximately 29.2 meters away horizontally when it hits the ground.
To find the horizontal distance traveled by the projectile when it hits the ground, we can use the time of flight and the horizontal component of the initial velocity.
The time of flight can be determined using the vertical motion of the projectile. We can use the equation:
h = (1/2) * g * t^2
where h is the initial height (15.4 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight.
Solving for t, we get:
t = sqrt((2 * h) / g)
Substituting the given values, we have:
t = sqrt((2 * 15.4) / 9.8) ≈ 1.75 seconds
Now, we can calculate the horizontal distance traveled using the formula:
distance = velocity * time
where velocity is the horizontal component of the initial velocity (16.7 m/s) and time is the time of flight (1.75 seconds).
Therefore, the horizontal distance traveled by the projectile when it hits the ground is:
distance = 16.7 * 1.75 ≈ 29.2 meters
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A power distribution system transmits 240MW of power at current of 500 A. At what voltage is the power distributed? Give your answer in kV. Show your full work 1 point for the steps 1 point for the correct value of voltage 1 point for the answer in kV
Therefore, the power is distributed at a voltage of 480 kV. The correct answer is 480 kv
In a power distribution system, power is transmitted from a power source to the consumers, who use it to perform work. A system can consist of a single source and one consumer or many sources and many consumers.
The power distribution system that we will consider in this problem transmits 240MW of power at a current of 500 A. We can use the following formula to calculate the voltage at which power is distributed:
V = P / I
whereV is the voltage in volts
P is the power in wattsI is the current in amperes
We can convert the power from megawatts to watts by multiplying by 10^6, so:
[tex]P = 240 MW * 10^6[/tex]
[tex]= 240 * 10^6 W[/tex]
Now we can substitute the given values and solve for V:
V = P / I
[tex]= (240 * 10^6 W) / 500 A[/tex]
= 480,000 V
To express the voltage in kilovolts, we divide by 1000:
V = 480,000 V / 1000
= 480 kV
Therefore, the power is distributed at a voltage of 480 kV.
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Consider the two line charges shown below. Both lines are the same length, L. The charge on the lines is the same magnitude, Q. One line charge starts at the origin, extends horizontally to the right (the positive x directionl, and has a positive charge, +Q. The other line charge starts at the origin, extends vertically downwards (the negative y direction) and has a negative charge, - Q. The point marked with an " X " is the same distance, L/2. from the middle of both line charges. This is the point x=L/2,y=−L/2. Q=0.59nC,L=0.72 m.(1nC=10−7C) What is the magnitude of the total electric feid at the point X due to these two line charges? Give your answer in N/C to at least three slgnifucant digits to avoid being counted off due to rounding. Do not include units in your ans
The magnitude of the total electric field at point X due to the two line charges is approximately 4.09 N/C.
To calculate the magnitude of the total electric field at point X due to the two line charges, we can use the principle of superposition. The electric field from each line charge can be calculated separately and then added together.
The electric field due to a line charge is given by the equation:
E = (k * λ) / r
Where E is the electric field, k is the Coulomb's constant (k = 9 * 10^9 N m^2/C^2), λ is the charge per unit length (Q/L), and r is the distance from the line charge.
Let's calculate the electric field from each line charge separately:
For the horizontal line charge:
λ = Q / L = 0.59nC / 0.72m = 0.819nC/m
r1 = L/2 = 0.72m / 2 = 0.36m
E1 = (9 * 10^9 N m^2/C^2) * (0.819nC/m) / 0.36m
For the vertical line charge:
λ = -Q / L = -0.59nC / 0.72m = -0.819nC/m (since it has a negative charge)
r2 = L/2 = 0.72m / 2 = 0.36m
E2 = (9 * 10^9 N m^2/C^2) * (-0.819nC/m) / 0.36m
Now, we can calculate the magnitude of the total electric field at point X by adding the contributions from both line charges:
|E_total| = |E1 + E2|
Substituting the calculated values:
|E_total| = |E1| + |E2|
To calculate the value of |E_total|, let's substitute the values we have obtained for E1 and E2:
E1 = (9 * 10^9 N m^2/C^2) * (0.819nC/m) / 0.36m ≈ 2.045 N/C
E2 = (9 * 10^9 N m^2/C^2) * (-0.819nC/m) / 0.36m ≈ -2.045 N/C
|E_total| = |E1| + |E2| = |2.045 N/C| + |-2.045 N/C| = 4.09 N/C
Therefore, the magnitude of the total electric field at point X due to the two line charges is approximately 4.09 N/C.
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What electric field, E, is necessary to drive a current of I=10 A through a silver wire with circular cross section of diameter d=1 mm ? Hint: The resistivity of silver is rho=1.59×10 −8Ω⋅m and is related to the conductivity σ by rho= σ1 what electric field is necessary to drive a current?
To determine the electric field necessary to drive a current through a silver wire, we can utilize Ohm's Law, which states that the current (I) flowing through a conductor is directly proportional to the electric field (E) and inversely proportional to the resistance (R) of the conductor.
The resistance of a conductor can be calculated using the formula:
R = (rho * L) / A
Where:
R is the resistance,
rho is the resistivity of the material,
L is the length of the conductor, and
A is the cross-sectional area of the conductor.
In this case, we are given the diameter of the silver wire (d), so we can calculate the cross-sectional area (A) using the formula:
A = (pi * d^2) / 4
Substituting the values into the formula, we have:
A = (pi * (1 mm)^2) / 4
Next, we can rearrange Ohm's Law to solve for the electric field (E):
E = I / (rho * L / A)
Given:
I = 10 A (current)
rho = 1.59 × 10^(-8) Ω⋅m (resistivity of silver)
d = 1 mm (diameter)
We need additional information to calculate the length of the conductor (L) in order to determine the electric field. Once the length is known, we can substitute all the values into the equation to find the electric field required to drive the given current through the silver wire.
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A 4200 Ω resistor is connected across a charged 0.800 nF capacitor. The initial current through the resistor, just after the connection is made, is measured to be 0.245 A .
1. What magnitude of charge was initially on each plate of this capacitor?
2. How long after the connection is made will it take before the charge is reduced to 1e of its maximum value?
1) The magnitude of charge initially on each plate of the capacitor is 0.196823.2 nC. 2) It will take approximately 5.28 ms for the charge to be reduced to 1e (10%) of its maximum value.
1. For finding the initial charge on each plate of the capacitor, use the formula for the charge stored in a capacitor:
Q = CV,
where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.
In this case, the voltage across the capacitor is given by Ohm's Law as:
V = IR,
where I is the current and R is the resistance.
Substituting the given values:
V = (0.245 A)(4200 Ω) = 1029 V.
Now, substituting this voltage and the capacitance into the charge formula:
Q = (0.800 nF)(1029 V) = 823.2 nC.
2. For determining the time it takes for the charge to reduce to one-tenth of its maximum value, use the formula for the discharge of a capacitor through a resistor:
[tex]t = RC ln(V_0/V)[/tex]
where t is the time, R is the resistance, C is the capacitance, [tex]V_0[/tex] is the initial voltage, and V is the final voltage (in this case, one-tenth of the initial voltage).
Plugging in the values:
t = (4200 Ω)(0.800 nF) ln(1029 V / (1029 V / 10)) = 5.28 ms.
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Transcribed image text:
A wall clock has a second hand 15.0 cm long. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Fast car, flat curve. What is the radial acceleration of the tip of this hand? Express your answer in centimeters per second squared.
The length of the second hand of the clock = 15.0 cmFormula used: The formula for calculating radial acceleration is given as ar = v^2 / r where ar is radial acceleration, v is the linear speed, and r is the radius of the circle.
Steps to solve the problem: Firstly, we need to calculate the linear speed of the second hand of the clock. The formula for calculating linear speed is given as v = ωr where ω is the angular velocity of the second hand of the clock and r is the radius of the circle. For a clock, we know that the second hand completes one revolution in one minute = 60 seconds. So, the angular velocity is given as:ω = 2π / t where t is the time taken for one revolution. So, t = 60 secondsTherefore,ω = 2π / t = 2π / 60 radians per second= π / 30 radians per second.The radius of the circle is given as r = 15.0 cmNow, we can calculate the linear speed:v = ωr = π / 30 × 15.0 cm/s = 0.5π cm/s .The radial acceleration is given as ar = v^2 / rar = (0.5π cm/s)^2 / 15.0 cm= (π^2) / (120 cm/s^2) = 0.082 cm/s^2
Answer: The radial acceleration of the tip of the second hand of the clock is 0.082 cm/s^2.
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An infinite line of charge produces a field of magnitude \( 3.4 \times 10^{4} \mathrm{~N} / \mathrm{C} \) at a distance of \( 2.5 \mathrm{~m} \). Calculate the linear charge density. Number Units
The linear charge density is 4.14 × 10-6 C/m. The linear charge density is 4.14 × 10^-6 C/m.
Given: An infinite line of charge produces a field of magnitude 3.4×104 N/C at a distance of 2.5 m.
To find: Calculate the linear charge density.
Solution: Let's consider the expression for the electric field due to an infinite line of charge is [tex]E = λ / 2πε₀r[/tex]
Where,λ is the linear charge densityε₀ is the permittivity of free space. r is the distance from the line of charge
The expression for the electric field at a distance of 2.5 m is
E = 3.4 × 104 N/C
The distance from the line of charge is r = 2.5 m
The expression for the electric field is [tex]E = λ / 2πε₀r[/tex]
We can find λ from the above expression [tex]λ = 2πε₀rE[/tex]
On substituting the known values, we get;
λ = 2 × π × 8.85 × 10-12 C² / N · m² × 2.5 m / (3.4 × 104 N/C)
= 4.14 × 10-6 C/m
∴ The linear charge density is 4.14 × 10-6 C/m.
The linear charge density is 4.14 × 10^-6 C/m.
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Find the electric field , E , between charge distributions 1.6 x 10³ meter apart and having a potential difference ( voltage difference ) of 240 volts .
The electric field between the charge distributions is 150 N/C.
Given that,
The distance between the charges, d = 1.6 x 10³ meters
Potential difference, V = 240 volts
The formula to calculate the electric field is
E = V/d = 240/1.6 x 10³= 150 N/C
Therefore, the electric field between the charge distributions is 150 N/C.
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Light from a xenon lamp with a wavelength of 553 nm illuminates two narrow slits. The spacing between two consecutive bright fringes is 8 mm on a screen that is 247 cm away. What is the spacing between the two slits in mm ?
To determine the spacing between the two slits, we can use the equation for the fringe spacing in a double-slit interference pattern. The spacing between the two slits is 8 mm.
The fringe spacing in a double-slit interference pattern is given by the equation d × sin(θ) = m × λ, where d is the spacing between the slits, θ is the angle of the bright fringe, m is the order of the fringe, and λ is the wavelength of the light.
In this case, the wavelength of the light is given as 553 nm, which is equivalent to 553 × 10⁻⁹ meters. The spacing between two consecutive bright fringes on the screen is given as 8 mm, which is equivalent to 8 × 10⁻³ meters.
The distance from the screen to the slits is given as 247 cm, which is equivalent to 247 × 10⁻² meters.
To find the spacing between the two slits, we rearrange the equation as d = (m × λ) / sin(θ). Since the angle θ is small, we can approximate sin(θ) as θ.
Substituting the given values, we have d = (m × 553 × 10⁻⁹ meters) / (8 × 10⁻³ meters / 247 × 10⁻² meters). Simplifying the expression, we find d = (m × 553 × 247) / 8 mm.
Therefore, the spacing between the two slits is (m × 553 × 247) / 8 mm.
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In a simple supported beam , write the moment equation of the beam shown in this video if it is loaded with 10 kN/m uniformly distributed over the entire simple span of 4 meters and a concentrated load of 20 kN applied 2 meters from the left support. Present your answer using the GAPSA form by indicating all the given data, what is being asked in the problem, the appropriate formula, the entire solutions / calculations and the answer accompanied with corresponding unit of measurement.
Thank you.
The moment equation of the beam is given by;∑M = 100 kN.m when x = 2 m.∑M = 240 kN.m when x = 8 m.
Given data: Length of the beam (L) = 4 meters. The uniformly distributed load = 10 kN/m, The concentrated load = 20 kN. The distance of the concentrated load from the left end of the beam = 2 meters GAPSA for: Given:Length of the beam (L) = 4 meters. The uniformly distributed load = 10 kN/m. The concentrated load = 20 kN. The distance of the concentrated load from the left end of the beam = 2 meters. Appropriate formula:The moment equation of the beam is given by;∑M = 0Where; ∑M = the summation of all moments acting on the beam.
Calculations: To find the moment equation of the beam, we need to calculate the reactions at the supports of the beam.∑Fy = 0Ay + By - 10 = 0 ...(1)∑M = 0Take moments about support B.Ay × 4 - 10 × 2 - 20 × 2 = 0Ay × 4 = 60Ay = 15 kNBy = 10 - AyBy = 10 - 15By = -5 kN (upward)The reactions at the supports of the beam are;Ay = 15 kN (downward)By = -5 kN (upward)Taking moments about support A;MA = 0Ay × 0 - 10 × 22 / 2 - 20 × 2 = 0MA = 40 kN.m. Taking moments about support B;MB = 0- By × 4 + 10 × 22 / 2 = 0MB = 30 kN.m. The moment equation of the beam is given by;∑M = 0MA - wx² / 2 - wLx / 2 - P (L - x) = 0Where;w = 10 kN/mP = 20 kNL = 4 meters.
Putting the values of MA, w, P, and L in the above equation;40 - 10x² / 2 - 10 × 4x / 2 - 20 (4 - x) = 0
Simplifying the above equation;10x² - 40x - 160 = 0x² - 4x - 16 = 0(x - 2) (x - 8) = 0x = 2 m or 8 m. When x = 2 m, the moment equation is;∑M = 0MA - wx² / 2 - wLx / 2 - P (L - x)= 0MA - 10 × 2² / 2 - 10 × 4 × 2 / 2 - 20 (4 - 2)= 0MA - 20 - 40 - 40 = 0MA = 100 kN.m. The moment equation of the beam when x = 2 m is;∑M = 100 kN.m. When x = 8 m, the moment equation is;∑M = 0MA - wx² / 2 - wLx / 2 - P (L - x)= 0MA - 10 × 8² / 2 - 10 × 4 × 8 / 2 - 20 (4 - 8)= 0MA - 160 - 160 + 80 = 0MA = 240 kN.m. The moment equation of the beam when x = 8 m is;∑M = 240 kN.m.
Therefore, the moment equation of the beam is given by;∑M = 100 kN.m when x = 2 m.∑M = 240 kN.m when x = 8 m
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Sius? What is the density of mercury at 100°C if that at 0°C is 13.6- if y Mercury 1.8×104 K-¹
The density of mercury at 100 °C, given that the density of mercury is 13.6 gcm⁻³ at 0 °C, is 13.4 g/cm³
How do i determine the of mercury at 100 °C?First, we shall obtain the new volume of the mercury. Details below:
Density at 0 °C (d₁) = 13.6 gcm⁻³
Mass (m) = 13.6 gInitial volume (V₁) = m/d₁ = 13.6 / 13.6 = 1 cm³Initial temperature (T₁) = 0 °CNew temperature (T₂) = 100 °CChange in temperature (ΔT) = 100 - 0 = 100 °CCoefficient of cubic expansion (α) = 1.8×10⁻⁴ K⁻¹New volume (V₂) = ?V₂ = V₁ ( 1 + αΔT)
= 1 × [1 + (1.8×10⁻⁴ × 100)]
= 1 × [1 + 0.018]
= 1.018 cm³
Finally, we shall obtain the density at 100°C. Details below:
New volume = 1.018 cm³ Mass of mercury = 13.6 gDensity at 100°C = ?Density at 100 °C = mass / new volume
= 13.6 / 1.018
= 13.4 g/cm³
Thus, the density of mercury at at 100 °C is 13.4 g/cm³
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Complete question:
What is the density of mercury at 100°C if that at 0°C is 13.6 gcm⁻³ and coefficient of cubic expansion of Mercury 1.8×10⁻⁴ K⁻¹
Find and expression for Electric field formed by a quarter of a charged ring with electric charge of the Q and with radius of R at the center.
The expression for the electric field formed by a quarter of a charged ring at the center is : [tex]E = (k * Q / (4\pi$r^2))[/tex]
To find the expression for the electric field formed by a quarter of a charged ring, we can consider the electric field contributions from each infinitesimally small charge element dQ on the ring.
Let's consider a small charge element on the ring at an angle θ from the positive x-axis.
The electric field dE produced by this charge element can be calculated using Coulomb's law as[tex]dE = (k * dQ) / r^2[/tex], where k is the electrostatic constant and r is the distance between the charge element and the point where we want to calculate the electric field (in this case, the center).
Since we are only considering a quarter of the ring, the total electric field at the center is obtained by integrating the electric field contributions from all the charge elements over the quarter ring.
The integration is performed with respect to θ, ranging from 0 to π/2 (quarter of the entire ring).
The resulting expression for the electric field E formed by a quarter of the charged ring at the center is:
[tex]E = \int\limits {[0 to$ \pi$/2] (k * dQ * cos(\theta)) / r^2} \,[/tex]
To evaluate the integral, we need to express dQ in terms of θ and relate it to the total charge Q of the ring.
The charge element dQ can be expressed as [tex]dQ = (Q / (4\pi$R)) * R * d\theta[/tex], where R is the radius of the ring and dθ is the differential angle.
Substituting this expression into the integral, we have:
[tex]E = \int\limits{[0 to$ $\pi$/2] (k * Q * cos(\theta)) / (4\pi$R * r^2) * R *} \, d\theta[/tex]
Simplifying further, we get:
[tex]E = (k * Q / (4\pi$$R)) *\int\limits {[0 to$ $\pi$/2] cos(\theta) / r^2 * R * } \, d\theta[/tex]
The integral of cos(θ) over the given range simplifies to sin(π/2) - sin(0), resulting in:
[tex]E = (k * Q / (4\pi$R)) * (1 - 0) / r^2 * R[/tex]
Finally, the expression for the electric field formed by a quarter of a charged ring at the center is:
[tex]E = (k * Q / (4\pi$r^2))[/tex]
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I need to run from point A to point C, via point B, for a total travel distance of ΔL. As pictured above, the distance between A and B is ΔL
1
, and the distance between B and C is ΔL
2
. I have time Δt to get from A to C, so I figure I can make it if I average v
avg
=ΔL/Δt for the entire run. Unfortunately, the ground between points A and B is covered in metal spikes, so I can only average a running speed v
1
for the first leg of my trip. a) ∗∗ What must my speed be during the second part of my trip, in order to average v
avg
for the whole trip? Write an equation expressing v
2
(the speed I need to average going from B to C ) in terms of the other symbols given (v
1
,ΔL
1
,ΔL
2
, and Δt). b) I need to run from point A to point C, via point B. The distance between A and B is 2.5 miles, and the distance between B and C is 7.5 miles. I have two hours to get from A to C, so I figure I can make it if I average 5mph for the entire run. Unfortunately, the ground between points A and B is covered in metal spikes, so I can only average a running speed of 2.5mph. What must my speed be during the second part of my trip in order to average 5mph for the whole trip?
a) The expression for v2 is given by: v2 = ΔL2 / (Δt - ΔL1 / v1)
b) The speed required in the second leg of the trip is 15 mph.
(a) We know that average speed is defined as the ratio of the total distance to the total time taken, i.e., v = ΔL/Δt. Here, the total distance is the sum of the distances of both legs. Thus, ΔL = ΔL1 + ΔL2. Let v2 be the speed in the second leg, then the time taken in the second leg is ΔL2/v2. The total time taken for the trip is given as Δt. Therefore,
ΔL1/v1 + ΔL2/v2 = Δt.
Let's isolate v2 in this expression:
ΔL2/v2 = Δt - ΔL1/v1
v2 = ΔL2 / (Δt - ΔL1 / v1)
(b) We can use the expression we derived in part (a) to calculate the speed required in the second leg. Let's substitute the given values: ΔL1 = 2.5 miles, ΔL2 = 7.5 miles, v1 = 2.5 mph, Δt = 2 hours, and v = 5 mph.
ΔL = ΔL1 + ΔL2 = 10 miles.
v2 = ΔL2 / (Δt - ΔL1 / v1)
= 7.5 / (2 - 2.5 / 2.5)
≈ 15 mph.
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A softball is hit so that it travels straight upward after being struck by a bat. An observer sees that the ball requires 2.10 s to reach its maximum height above the ground, 73.5 feet. (a) At what height above the ground was the ball struck, and (b) what was the ball’s initial velocity? An
(a) The height above the ground at which the ball was struck is 140.6 feet.
(b) The initial velocity of the ball is 55.6 ft/s.
Here, you are given that a softball is hit so that it travels straight upward after being struck by a bat and the observer sees that the ball requires 2.10 s to reach its maximum height above the ground, 73.5 feet.
(a) To calculate at what height above the ground was the ball struck, we will use the formula below:
Δy = Vit + (1/2)at², where
Δy = 73.5 feet, Vi = ?, a = -32.2 ft/s² (since the acceleration due to gravity is opposite to the direction of motion of the ball), and t = 2.10 s.
Substituting all the given values in the above formula, we get
73.5 = Vi(2.10) + (1/2)(-32.2)(2.10)²
Simplifying the equation, we get
73.5 = 2.205Vi - 22.959
Multiplying throughout by 1000, we get
73500 = 2205Vi - 22959
Rearranging the terms, we get
2205Vi = 96259
Dividing throughout by 2205, we get
Vi = 96259/2205
Vi = 43.6 ft/s
Now, to find at what height above the ground was the ball struck, we will use the above equation again.
Δy = Vit + (1/2)at²
Substituting all the given values, we get
Δy = 43.6(2.10) + (1/2)(-32.2)(2.10)²
Δy = 140.6 feet
Therefore, at what height above the ground was the ball struck = 140.6 feet.
(b) To find the ball's initial velocity, we can use the same formula, i.e.
Δy = Vit + (1/2)at²
Substituting the given values, we get
73.5 = Vi(2.10) + (1/2)(-32.2)(2.10)²
Rearranging the terms, we get
Vi = (73.5 - (1/2)(-32.2)(2.10)²)/2.10
Vi = 55.6 ft/s
Therefore, the ball’s initial velocity = 55.6 ft/s.
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A rocket is launched at an angle of 52 degree above the horizontal with an initial speed of 101 m/s. The rocket moves for 3s along with its initial line of motion with an acceleration of 31 m/s2. At this time, its engines fail and the rocket proceeds to move as a projectile. (a) Find the maximum altitude reached by the rocket. m (b) Find its total time of flight. s (c) Find its horizontal range. m
The answer is a) the maximum altitude reached by the rocket is 313.18 m, b) the total time of flight of the rocket is 7.71 s, c) the horizontal range of the rocket is 479.5 m. Given, Angle of the rocket, θ = 52°; Initial speed of the rocket, u = 101 m/s; Initial acceleration, a = 31 m/s²; Time, t = 3 s
(a) When the rocket moves with an initial acceleration of 31 m/s² for 3s along with its initial line of motion, Then, initial velocity, v = u + at= 101 + (31 × 3)= 101 + 93= 194 m/s
Let, the maximum height reached by the rocket be h m. Then, at maximum height, vertical velocity of the rocket, v = 0 m/s
Initial vertical velocity of the rocket, u₀ = usinθ= 101sin52°= 79.12 m/s
Using the third equation of motion,v² = u₀² + 2gh0 = u₀² / 2g= 79.12² / (2 × 9.8)= 313.18 m
Therefore, the maximum altitude reached by the rocket is 313.18 m.
(b) When the rocket becomes a projectile, there is no vertical acceleration.So, acceleration, a = 0 m/s² Let the total time of flight of the rocket be T.Using the first equation of motion,Total horizontal distance traveled by the rocket,
R = ucosθ × T= 101cos52° × T= 101 × 0.614 × T= 62.214T
Using the second equation of motion,Vertical displacement of the rocket, h = u₀T + 1/2 × a × T²= 79.12T + 1/2 × 0 × T²= 79.12T
Using the third equation of motion,v² = u² + 2ghh = 0 + 2 × 9.8 × h= 19.6hT = (2h / g)½= (2 × 19.6h / 9.8)½= (4h / g)½
Using the value of h obtained above,Total time of flight of the rocket,T = (4 × 313.18 / 9.8)½= 7.71 s
Therefore, the total time of flight of the rocket is 7.71 s.
(c) From the equation,R = ucosθ × T= 101cos52° × T= 62.214T
Using the value of T obtained above,Horizontal range of the rocket, R = 62.214 × 7.71= 479.5 m
Therefore, the horizontal range of the rocket is 479.5 m.
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A cheetah can accelerate from rest to 25.0 m/s in 6.22 s. Assuming constant acceleration, (a) how far has the cheetah run in this time? (b) After sprinting for just 3.11 s, is the cheetah’s speed 12.5 m/s, greater than 12.5 m/s, or less than 12.5 m/s? Explain. (c) What is the cheetah’s average speed for the first 3.11 s of its sprint? For the second 3.11 s of its sprint? (d) Calculate the distance covered by the cheetah in the first 3.11 s and the second 3.11 s.
The cheetah covers a distance of approximately 19.4 m in the first 3.11 s and 58.1 m in the second 3.11 s.
(a) To calculate the distance the cheetah has run, we can use the formula:
distance = (initial velocity * time) + (0.5 * acceleration * time^2).
Given that the initial velocity is 0 m/s (since the cheetah starts from rest), the final velocity is 25.0 m/s, and the time is 6.22 s, we can substitute these values into the equation:
distance = (0 * 6.22) + (0.5 * acceleration * 6.22^2).
We need to solve for the acceleration first. Rearranging the equation for acceleration:
acceleration = (final velocity - initial velocity) / time.
acceleration = (25.0 m/s - 0 m/s) / 6.22 s
= 4.02 m/s^2.
Substituting the acceleration into the distance equation:
distance = (0.5 * 4.02 m/s^2 * (6.22 s)^2)
= 77.5 m.
Therefore, the cheetah has run a distance of 77.5 meters in this time.
(b) After sprinting for just 3.11 s, we can determine the cheetah's speed by using the formula:
final velocity = initial velocity + (acceleration * time).
Given that the initial velocity is 0 m/s and the time is 3.11 s, we can substitute these values into the equation:
final velocity = 0 m/s + (4.02 m/s^2 * 3.11 s)
= 12.5 m/s.
The cheetah's speed after sprinting for 3.11 s is exactly 12.5 m/s.
(c) The cheetah's average speed for the first 3.11 s of its sprint is the distance covered divided by the time taken. We can use the formula:
average speed = distance / time.
Using the same time of 3.11 s, we can calculate the average speed for the first part of the sprint:
average speed = 77.5 m / 3.11 s ≈ 24.9 m/s.
For the second 3.11 s of its sprint, the average speed would be the same, as the acceleration is constant throughout the entire sprint.
(d) The distance covered by the cheetah in the first 3.11 s and the second 3.11 s can be calculated using the formula:
distance = (initial velocity * time) + (0.5 * acceleration * time^2).
For the first 3.11 s:
distance = (0 m/s * 3.11 s) + (0.5 * 4.02 m/s^2 * (3.11 s)^2) ≈ 19.4 m.
For the second 3.11 s:
distance = (12.5 m/s * 3.11 s) + (0.5 * 4.02 m/s^2 * (3.11 s)^2) ≈ 58.1 m.
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What is the speed of an electron that has been accelerated from rest through a potential difference of 3400 V? (express your answer in m/s using scientific notation, e.g. 3.84*10^7)
The speed of an electron that has been accelerated from rest through a potential difference of 3400 V is 1.08 × 10⁵ m/s.
The equation used for calculating the speed of an electron is given by the following:
Given that the potential difference is 3400 V.
Therefore, we can use the equation above to find the speed of the electron.
v = √((2*e*V)/m)
Where: v is the velocity of an electron, e is the electron's charge V is the potential difference between the plates, and m is the mass of an electron.
Substitute the given values of e, V and m in the above equation.
v = √((2*1.6*10⁻¹⁹*3400)/9.11*10⁻³¹)
1.6 * 10⁻¹⁹ is the charge of an electron 9.11 * 10⁻³¹ is the mass of an electron.
v = √(0.01172)
Converting the value to scientific notation:
v = 1.08*10⁵ m/s (rounded off to two significant figures)
Thus, The speed of an electron that has been accelerated from rest through a potential difference of 3400 V is 1.08 × 10⁵ m/s.
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A catapult launches a rock at an initial angle of θ=60 degrees. What is the minimum initial velocity v
i
of the rock so that it will land on a platform that is 35 m away and 10 m tall?
There is no minimum initial velocity (vᵢ) required for the rock to land on the platform.
find the minimum initial velocity (vᵢ) of the rock, we can analyze the projectile motion of the rock.
The horizontal motion and vertical motion are independent of each other.
Horizontal Motion:
The horizontal displacement (range) is given as 35 m.
Vertical Motion:
The vertical displacement (height) is given as 10 m.
The initial vertical velocity (vᵢ⋅sinθ) is unknown.
The acceleration due to gravity (g) is approximately 9.8 m/s².
The final vertical velocity (v_f) is 0 m/s at the highest point.
Using the kinematic equation for vertical motion:
v_f² = vᵢ² + 2gΔy
At the highest point, v_f = 0, so the equation becomes:
0 = vᵢ² + 2gΔy
Substituting the known values, we have:
0 = vᵢ² + 2(9.8 m/s²)(10 m)
Simplifying the equation, we find:
0 = vᵢ² + 196 m²/s²
Now, let's analyze the horizontal motion:
The horizontal velocity (vᵢ⋅cosθ) remains constant throughout the motion.
The horizontal displacement (range) is given as 35 m.
Using the equation for horizontal motion:
range = vᵢ⋅cosθ⋅time
Since the range is given as 35 m, we can rearrange the equation to solve for time:
time = range / (vᵢ⋅cosθ)
Substituting the known values, we have:
time = 35 m / (vᵢ⋅cos60°)
Simplifying the equation, we find:
time = 35 m / (vᵢ⋅0.5)
time = 70 m / vᵢ
combine the vertical and horizontal motion equations:
We know that the total time of flight is the same for both the horizontal and vertical motion.
we can equate the time derived from vertical motion (70 m / vᵢ) to the time derived from horizontal motion.
70 m / vᵢ = time = range / (vᵢ⋅cos60°)
Substituting the known values, we have:
70 m / vᵢ = 35 m / (vᵢ⋅cos60°)
Simplifying the equation, we find:
2 = cos60°
Therefore, cos60° = 0.5.
Substituting this value into the equation, we have:
70 m / vᵢ = 35 m / (vᵢ⋅0.5)
Simplifying the equation, we find:
70 m / vᵢ = 70 m / vᵢ
This equation holds true regardless of the value of vᵢ.
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A rescue helicopter is lifting a man (weight \( =623.215888270436 \mathrm{~N} \) ) from a capsized boat by means of a cable and harness. (a) What is the tension in the cable when the man is given an i
The tension in the cable when the man is given an initial upward acceleration of 3.8 m/s² can be calculated using the equation[tex]\(T = mg + ma\[/tex] .
Given:
Weight of the man, [tex]\(mg = 623.215888270436 \, \text{N}\)[/tex]
Initial upward acceleration, [tex]\(a = 3.8 \, \text{m/s}^2\)[/tex]
To find the tension in the cable, we need to consider both the weight of the man and the additional force required to accelerate him upward.
First, we calculate the force due to gravity (weight):[tex]\(mg = 623.215888270436 \, \text{N}\).[/tex]
Next, we calculate the additional force required to accelerate the man upward:[tex]\(ma = 623.215888270436 \, \text{N} \times 3.8 \, \text{m/s}^2\)[/tex].
Finally, we add these two forces together to obtain the tension in the cable:[tex]\(T = mg + ma\).[/tex]
By substituting the values into the equation, you can calculate the tension in the cable when the man is given an initial upward acceleration.
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A 1100-kg car pulls a boat on a trailer. (Enter the magnitude.)
(a) What total force (in N) resists the motion of the car, boat, and trailer, if the car exerts a 1880-N force on the road and produces an acceleration of 0.570 m/s2? The mass of the boat plus trailer is 660 kg.
(b)
What is the force (in N) in the hitch between the car and the trailer if 80% of the resisting forces are experienced by the boat and trailer?
(a) To determine the total force resisting the motion of the car, boat, and trailer, we need to consider the forces acting on the system.
The force exerted by the car on the road, which propels the entire system, is given as 1880 N. The mass of the car is 1100 kg, and it produces an acceleration of 0.570 m/s^2.
Using Newton's second law (F = ma), we can find the net force acting on the system:
F_net = ma = (1100 kg)(0.570 m/s^2) = 627 N
Therefore, the net force acting on the system is 627 N.
To find the total force resisting the motion, we subtract the force exerted by the car on the road from the net force:
Total force resisting motion = F_net - Force exerted by car on road
= 627 N - 1880 N
= -1253 N
The negative sign indicates that the force is opposing the motion of the system.
Therefore, the magnitude of the total force resisting the motion of the car, boat, and trailer is 1253 N.
(b) If 80% of the resisting forces are experienced by the boat and trailer, then the remaining 20% of the resisting force is experienced by the car.
Let's denote the force in the hitch between the car and the trailer as F_hitch.
If 80% of the resisting forces are experienced by the boat and trailer, then the force in the hitch can be calculated as follows:
F_hitch = 20% of the total force resisting motion
Since we found (motion to be 1253 N, we can calculate the force in the hitch:
F_hitch = 20% of 1253 N
= (20/100) * 1253 N
= 0.20 * 1253 N
= 250.6 N
Therefore, the force in the hitch between the car and the trailer is approximately 250.6 N.
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A particle (q=+3.5μC) is released from rest when it is 2.0 m from a charged particle which is held at rest. After the positively charged particle has moved 1.0 m toward the fixed particle, it has a kinetic energy of 35 mJ. What is the charge (in μC ) on the fixed particle?
The charge on the fixed particle is approximately 3.334 μC. To solve this problem, we can use the principles of conservation of energy and Coulomb's law.
Charge of the released particle, q = +3.5 μC
Distance between the particles initially, r1 = 2.0 m
Distance the released particle moves, r2 = 1.0 m
Kinetic energy of the released particle, KE = 35 mJ = 35 × 10^-3 J
We can determine the work done by the electrostatic force as the released particle moves from distance r1 to r2. The work done is equal to the change in potential energy.
The change in potential energy is given by:
ΔPE = PE2 - PE1
where PE2 is the potential energy of the system at distance r2 and PE1 is the potential energy of the system at distance r1.
The potential energy due to the electrostatic force between two charged particles is given by:
PE = k * (|q1 * q2|) / r
where k is the Coulomb's constant (8.99 × 10^9 N m^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between them.
At distance r1:
PE1 = k * (|q1 * q2|) / r1
At distance r2:
PE2 = k * (|q1 * q2|) / r2
The work done is then:
ΔPE = PE2 - PE1 = k * (|q1 * q2|) * (1/r2 - 1/r1)
Since the released particle is initially at rest, its initial total mechanical energy is zero. Therefore, the work done by the electrostatic force is equal to the change in kinetic energy:
ΔPE = KE
Substituting the given values:
k * (|q1 * q2|) * (1/r2 - 1/r1) = 35 × 10^-3
Solving for |q1 * q2|:
|q1 * q2| = (35 × 10^-3) * (r2 - r1) / k
Now, we can substitute the values and solve for the charge on the fixed particle:
|q1 * q2| = (35 × 10^-3) * (1.0 - 2.0) / (8.99 × 10^9)
|q1 * q2| = -1.167 × 10^-11 C^2
Since the charge of the released particle is q1 = +3.5 μC, we can find the charge of the fixed particle, q2:
|q1 * q2| = |q1| * |q2|
|q2| = |q1 * q2| / |q1|
|q2| = (-1.167 × 10^-11) / (3.5 × 10^-6)
|q2| ≈ -3.334 × 10^-6 C
Since charge cannot be negative, we take the magnitude of |q2|:
|q2| ≈ 3.334 μC
Therefore, the charge on the fixed particle is approximately 3.334 μC.
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