The amplitude of the oscillations is 6.5 centimeters, keeping three significant digits.
To determine the amplitude of the oscillations, we can use the equation for the displacement of a mass-spring system- x(t) = A * cos(ωt + φ)
where x(t) is the displacement as a function of time, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.
The angular frequency ω can be calculated using the formula:
ω = sqrt(k/m) where k is the stiffness of the spring and m is the mass.
Given that the mass is 0.73 kg and the stiffness is 31.1 N/m, we can calculate the angular frequency: ω = sqrt(31.1 / 0.73) ≈ 6.211 rad/s
To find the amplitude A, we need to consider the initial conditions. At t = 0, the displacement x(0) is 6.5 centimeters. Plugging these values into the equation, we have: 6.5 cm = A * cos(0 + φ)
Since the cosine function has a maximum value of 1, the amplitude A is equal to the initial displacement: A = 6.5 cm
Therefore, the amplitude of the oscillations is 6.5 centimeters, keeping three significant digits.
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closed surface encloses a net charge of 3.10μC. What is the net electric flux through the surface? Express your answer in newton-meters squared per coulomb. If the electric flux through a closed surface is determined to be 2.10 N⋅m
2
/C, how much charge is enclosed by the surface? Express your answer in coulombs.
The amount of charge enclosed by the surface is approximately 1.86 × 10⁻¹¹ C.
If a closed surface encloses a net charge of 3.10μC, the net electric flux through the surface will be given by;
ϕ=Q/ϵ₀
where ϕ = electric flux through a closed surface
Q = net charge enclosed by the surface
ϵ₀ = permittivity of free space
Therefore, the electric flux through a closed surface that encloses a net charge of 3.10μC will be given by;
ϕ = 3.10μC / 8.85 × 10⁻¹² C²/N m²
ϕ ≈ 3.50 × 10¹⁰ N m²/C
The net electric flux through the surface is 3.50 × 10¹⁰ N m²/C.
If the electric flux through a closed surface is determined to be 2.10 N⋅m²/C, the amount of charge enclosed by the surface will be given by;
ϕ=Q/ϵ₀
Q = ϕ × ϵ₀
Therefore, the charge enclosed by the surface will be given by;Q = 2.10 N⋅m²/C × 8.85 × 10⁻¹² C²/N m²Q ≈ 1.86 × 10⁻¹¹ C
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a) A cylinder with length 8 m and diameter 1.5 m is partially submerged and upright in water. The density of the cylinder is 90% that of water. Determine the amount of mass that can be added to the top of the cylinder before it is completely submerged. Take density of water as 1000 kg/m
3
. b) A spring mass system with mass, m=150 kg and spring constant, k=500kN/m, is stretched 4 cm from its equilibrium position and then released. Determine the i. the frequency and period of oscillation ii. the change in mass if the period increases by 25%
a) The amount of mass that can be added to the top of the cylinder before it is completely submerged is 5,059.65 kg.
b) The change in mass is 1.5625 kg.
a) Density of the cylinder = 90% of density of water = 90/100 * 1000 = 900 kg/m³Volume of the cylinder = πr²h where r = diameter/2 = 1.5/2 = 0.75m∴ Volume of cylinder = π(0.75)² × 8 = 14.137 m³Buoyant force = Volume of water displaced × Density of water
Volume of water displaced = volume of cylinder partially submerged
Volume of cylinder submerged = π(0.75)² × h = 14.137/2 = 7.0685 m³ where h is height of cylinder submerged
By definition of density, Mass of cylinder partially submerged/Volume of cylinder partially submerged = 900 kg/m³∴ Mass of cylinder partially submerged = 900 × 7.0685 = 6351.65 kg
We know that force acting on cylinder = Weight of cylinder partially submerged + weight of mass added to top of cylinder
Let the amount of mass that can be added to the top of cylinder before it is completely submerged be m kg. Then force acting on cylinder = 900 × g × Volume of cylinder partially submerged + (6351.65 + m) × g, where g is acceleration due to gravity.
Force acting on cylinder = Buoyant force acting on cylinder∴ 900 × g × Volume of cylinder partially submerged + (6351.65 + m) × g = 900 × g × Volume of cylinder submerged∴ m = 900 × g × (Volume of cylinder submerged - Volume of cylinder partially submerged)/g= 900 × (14.137/2 - 7.0685) = 5,059.65 kg
b) Given, m = 150 kg, k = 500 kN/m, x = 4 cm = 0.04 m,
i) The frequency of oscillation is given by,ω = √(k/m) where ω = 2πf∴ f = ω/2π= √(k/m)/(2π)= √(500 × 10³/150)/(2π)= 12.91 Hz
Period, T = 1/f = 1/12.91 = 0.0773 s
ii) Let the change in mass be dm kg.
Then the period of oscillation, T' = 2π√(m + dm)/k= 2π√(m + 0.25m)/k= 2π√(1.25m)/k= 2π/√k * √1.25m
∴ T'/T = √1.25m
∴ √1.25m = T'/T= 1 + 25/100 = 1.25
∴ m = (1.25)²/1.25 × 150= 1.5625 kg.
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A basebali thrown at an angle of 55.0
∘
above the horizontni strikes a bullding 180 m away at a point 7.00 m above the point from which it is thrown. Part A ignore ait resistance. Find the magnitude of the inisal velocity of the baseball (the velocry with which the baseball is thrown). Express your answer in meters per second. Part B Find the magnitude of the velocity of the basebal just before it atrikes the building. Express your answer in meters per second. Find the direction of the velocity of the baseball just before is strikes the builing. Express your answer in degrees.
The magnitude of the initial velocity of the baseball is approximately 32.1 m/s. We can analyze the horizontal and vertical components separately. The direction of the velocity just before the baseball strikes the building is approximately 52.1 degrees.
Part A: To find the magnitude of the initial velocity of the baseball, we can analyze the horizontal and vertical components separately.
Given:
Angle of projection (θ) = 55.0°
Horizontal distance (d) = 180 m
Vertical distance (h) = 7.00 m
The horizontal component of velocity (v_x) remains constant throughout the motion, so:
v_x = d / t
where t is the time of flight.
The vertical component of velocity (v_y) can be found using the equation:
h = v_y * t - (1/2) * g * t²
where g is the acceleration due to gravity.
We can also relate the initial velocity (v₀) to the horizontal and vertical components:
v_x = v₀ * cosθ
v_y = v₀ * sinθ
From the above equations, we can express t in terms of v₀ and solve for v₀:
d / (v₀ * cosθ) = h / (v₀ * sinθ) - (1/2) * g * (h / (v₀ * sinθ))²
Simplifying the equation, we get:
d * sinθ - (1/2) * g * (h / sinθ)² = v₀² * cosθ
Substituting the known values:
180 m * sin(55.0°) - (1/2) * 9.8 m/s² * (7.00 m / sin(55.0°))² = v₀² * cos(55.0°)
Solving this equation for v₀, we find:
v₀ ≈ 32.1 m/s
Therefore, the magnitude of the initial velocity of the baseball is approximately 32.1 m/s.
Part B: To find the magnitude of the velocity of the baseball just before it strikes the building, we can use the vertical component of velocity. Since there is no vertical acceleration (ignoring air resistance), the magnitude of the vertical velocity component remains the same as the initial velocity component:
v_y = v₀ * sinθ
Substituting the known values:
v_y = 32.1 m/s * sin(55.0°)
v_y ≈ 26.2 m/s
The magnitude of the velocity of the baseball just before it strikes the building is approximately 26.2 m/s.
To find the direction of the velocity just before it strikes the building, we can use the horizontal component of velocity. The horizontal component remains constant, so:
v_x = v₀ * cosθ
Substituting the known values:
v_x = 32.1 m/s * cos(55.0°)
v_x ≈ 20.9 m/s
Therefore, the direction of the velocity just before the baseball strikes the building can be found using the inverse tangent:
θ = arctan(v_y / v_x)
θ ≈ arctan(26.2 m/s / 20.9 m/s)
θ ≈ 52.1°
The direction of the velocity just before the baseball strikes the building is approximately 52.1 degrees.
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A dipole is centered at the origin, and is composed of charged particles with charge +e and −e, separated by a distance 5×10
−10
m along the y axis. The +e charge is on the −y axis, and the −e charge is on the +y axis. Part 1 A proton is located at <0,2×10
−8
,0>m. What is the force on the proton, due to the dipole?
F
=< >N Attempts: 0 of 4 used
The force on the proton due to the dipole is F = 0.
A dipole is centered at the origin, and is composed of charged particles with charge +e and −e, separated by a distance 5×10−10m along the y axis.
The +e charge is on the −y axis, and the −e charge is on the +y axis.
The electric force (F) on a proton due to the dipole can be calculated as shown below:
[tex]F= F1 + F2F1 = (k*q1*q)/(r1)^2[/tex]
[tex]F2 = (k*q2*q)/(r2)^2[/tex]
In the given situation, both charges are equal and opposite in nature and situated at a fixed distance from each other. So, they will produce forces of equal magnitudes that cancel each other out.
So the net force on the proton is zero.
Hence, the force on the proton due to the dipole is F = 0.
Therefore, the answer is:< 0, 0, 0>N
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You apply Coulomb's law separately for charges +e and -e, calculate their individual forces on the proton which will be along opposite directions along y-axis. Subtract them to get the resultant force.
Explanation:To calculate the force on the proton due to the dipole, we use Coulomb's law, which is about the interaction between charged particles. Coulomb's Law states F = k*q1*q2)/(r^2) where F is the force, k is Coulomb's constant (9.0 x 10^9 N*m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
First, you find the forces on the proton due to charge +e and -e separately. The proton's charge will interact with the +e charge and -e charge individually with different distances. So, calculate the force due to each. Then, because the force due to each is in the opposite direction, subtract to find the total force.
Remember, since force is a vector, it has both magnitude and direction; you need to account for this in your calculations. All measurements of displacement should be along the y-axis, and any deviation from that should be evaluated by its y component.
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Two 3.0 cm-diameter disks tace each other, 1.9 mm apart. They are charged to ±19nC. What is the electric field strength between the disks ? Express your answer to two significant figures and include the appropriate units.
The electric field strength between the two charged disks can be calculated using the formula E = σ / (2 × ε0), where σ is the surface charge density and ε0 is the permittivity of free space. Given the charge on each disk (±19nC) and the distance between them (1.9 mm), the electric field strength is approximately 1.51 x 10⁶ N/C.
The electric field between the disks can be calculated using the formula:
E = σ / (2 × ε0)
where σ is the surface charge density, ε0 is the permittivity of free space, and E is the electric field strength.
The surface charge density can be calculated as:
σ = Q / A
where Q is the charge on each disk (±19nC in this case) and A is the area of each disk (πr², where r is the radius of the disk).
Given that the disks have a diameter of 3.0 cm, the radius is 1.5 cm or 0.015 m. Therefore, the area of each disk is:
A = πr² = π(0.015 m)² = 7.07 x 10⁻⁴ m²
The surface charge density is:
σ = Q / A = ±19nC / 7.07 x 10⁻⁴ m² = ±2.68 x 10⁻⁵ C/m²
Now, we can use the formula for electric field to find the value of E:
E = σ / (2 × ε0) = ±2.68 x 10⁻⁵ C/m² / (2 × 8.85 x 10⁻¹² F/m) = ±1.51 x 10⁶ N/C
Therefore, the electric field strength between the disks is approximately 1.51 x 10⁶ N/C.
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True or false : the conservation of momentum is most closely related to newton's
The statement is false. Conservation of momentum is most closely related to the first law of motion, which states that an object at rest will remain at rest or in motion at a constant velocity unless acted upon by an unbalanced force.
The statement "True or false: the conservation of momentum is most closely related to Newton's" is false.What is conservation of momentum?
The conservation of momentum principle states that if no external forces act on a system of objects, then their total momentum remains constant. In an isolated system, the total momentum of the system before the collision is equal to the total momentum after the collision.
The law of conservation of momentum is an essential law of physics that describes how the momentum of a system changes when it interacts with its surroundings. When the momentum of an object changes, it affects the momentum of everything around it.
The law of conservation of momentum is a fundamental principle that describes how the momentum of a system changes when it interacts with its surroundings.
It is based on the principle of inertia, which states that an object will remain at rest or continue moving at a constant velocity unless acted upon by an external force.
Thus, the statement is false. Conservation of momentum is most closely related to the first law of motion, which states that an object at rest will remain at rest or in motion at a constant velocity unless acted upon by an unbalanced force.
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The statement is false. Conservation of momentum is most closely related to the first law of motion, which states that an object at rest will remain at rest or in motion at a constant velocity unless acted upon by an unbalanced force.
The statement "True or false: the conservation of momentum is most closely related to Newton's" is false.What is conservation of momentum?
The conservation of momentum principle states that if no external forces act on a system of objects, then their total momentum remains constant. In an isolated system, the total momentum of the system before the collision is equal to the total momentum after the collision.
The law of conservation of momentum is an essential law of physics that describes how the momentum of a system changes when it interacts with its surroundings. When the momentum of an object changes, it affects the momentum of everything around it.
The law of conservation of momentum is a fundamental principle that describes how the momentum of a system changes when it interacts with its surroundings.
It is based on the principle of inertia, which states that an object will remain at rest or continue moving at a constant velocity unless acted upon by an external force.
Thus, the statement is false. Conservation of momentum is most closely related to the first law of motion, which states that an object at rest will remain at rest or in motion at a constant velocity unless acted upon by an unbalanced force.
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Signal x(t)=2cos(πt+
4
π
) is sampled at 10 Hz sampling rate. Represent output in discrete time sequence x[n]. Also, find the digital frequency ω in rad/sample.
To represent the signal x(t) in a discrete time sequence x[n], we need to sample the continuous signal at the given sampling rate of 10 Hz. The sampling rate determines how frequently we take samples of the continuous signal.
In this case, the continuous signal x(t) is given as 2cos(πt+4π). The general formula for sampling a continuous signal is x[n] = x(nT), where n represents the sample number and T is the sampling period (1/frequency).
The sampling period can be calculated as T = 1/f, where f is the sampling rate. In this case, f = 10 Hz, so T = 1/10 = 0.1 seconds.
To find the discrete time sequence x[n], we substitute nT into the original signal x(t). So, x[n] = 2cos(π(nT)+4π).
Since T = 0.1 seconds, the discrete time sequence becomes x[n] = 2cos(πn(0.1)+4π).
To find the digital frequency ω in rad/sample, we can use the relationship ω = 2πf, where f is the sampling rate. In this case, ω = 2π(10) = 20π rad/sample.
Therefore, the output in discrete time sequence x[n] is 2cos(πn(0.1)+4π), and the digital frequency ω is 20π rad/sample.
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Ceetert: torpat. Eapress your answer in newton fimes metert squared per codiont A point charge q1=4.05nC is located on the x-axis at x=1.85 m, and a second point charge q2 =−6.05nC is on the y-axis at y=1.10 m Express your answer in newton times meters squared per coulomb.
The force experienced by [tex]q_2[/tex] on [tex]q_1[/tex] is [tex]-7.45 x 10^-^7N[/tex]. The unit of the answer is N·m²/C.
For calculation of the force experienced by [tex]q_2[/tex] on [tex]q_1[/tex], we have to calculate the distance between [tex]q_1[/tex] and [tex]q_2[/tex] first. We can calculate it by using the Pythagorean theorem:
[tex]r^2 = x^2 + y^2r^2[/tex]
[tex]= (1.85)^2 + (1.10)^2[/tex]
[tex]= 2.19 m[/tex]
Now, we can use Coulomb's law to calculate the force:
[tex]F = k * q_1 * q_2 / r^2[/tex] where k = Coulomb's constant = [tex]9 x 10^9 Nm^2/C^2[/tex], [tex]q_1 = 4.05 nC[/tex] and [tex]q_2 = -6.05 nC[/tex]
[tex]F = (9 x 10^9 N m^2/C^2) * (4.05 x 10^-^9 C) * (-6.05 x 10^-^9C) / (2.19)^2[/tex]
[tex]F = -7.45 x 10^-^7 N[/tex]
The unit of the answer is N·m²/C.
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A ball of mass \( 0.400 \mathrm{~kg} \) rolls a distance of \( 35.6 \mathrm{~m} \) along a path that is angled at \( 8.00 \) degrees below the horizontal (it is rolling down a hill). The work done by
To calculate the work done by the ball, we need to determine the force acting on the ball and the displacement along the path.
The force acting on the ball can be divided into two components: the force of gravity and the force of friction. The force of gravity can be calculated using the formula:
[tex]\( F_{\text{gravity}} = m \cdot g \),[/tex]
where \( m \) is the mass of the ball and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).
The force of friction can be calculated using the formula:
[tex]\( F_{\text{friction}} = \mu \cdot F_{\text{normal}} \),[/tex]
where \( \mu \) is the coefficient of friction and \( F_{\text{normal}} \) is the normal force. The normal force can be determined by decomposing the weight of the ball into components perpendicular and parallel to the inclined path:
[tex]\( F_{\text{normal}} = m \cdot g \cdot \cos(\theta) \),[/tex]
where \( \theta \) is the angle of the incline (8 degrees in this case).
The coefficient of friction depends on the nature of the surfaces in contact. Since the ball is rolling without slipping, we can assume that the coefficient of rolling friction (\( \mu_{\text{rolling}} \)) is applicable. For most surfaces, this value ranges between 0.01 and 0.03.
Once we have the force acting on the ball, we can calculate the work done using the formula:
\( \text{Work} = \text{force} \cdot \text{displacement} \cdot \cos(\theta) \).
Now, let's calculate the work done step by step.
Given:
Mass of the ball (m) = 0.400 kg
Distance rolled (d) = 35.6 m
Angle of the incline (θ) = 8.00 degrees
Step 1: Calculate the force of gravity (F_gravity):
\( F_{\text{gravity}} = m \cdot g \)
\( F_{\text{gravity}} = 0.400 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)
Step 2: Calculate the normal force (F_normal):
\( F_{\text{normal}} = m \cdot g \cdot \cos(\theta) \)
\( F_{\text{normal}} = 0.400 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos(8.00^\circ) \)
Step 3: Calculate the force of friction (F_friction):
\( F_{\text{friction}} = \mu_{\text{rolling}} \cdot F_{\text{normal}} \)
Step 4: Calculate the work done (Work):
\( \text{Work} = F_{\text{friction}} \cdot d \cdot \cos(\theta) \)
By plugging in the values, you can calculate the work done by the ball along the inclined path.
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(a) What is the ate of energy loss in watts per square meter through a glass window 4.5 mm thick if the outside temperature is - roof and the inside temperature is +72
∘
F ? Wm
2
(b) A storm window having the jame thickneas of glass is installed parallel to the firet window, with an alr gap of 7.5 cin between the two Windows. What now is the rate of enorgy loos if conduction is the only important energytoss mechanism? Wrim'
The rate of energy loss through the glass window is about 262.5 W/m^2. We can use the formula: Q/t = kA(T2 - T1)/d. The rate of energy loss through the two glass panes with an air gap between them is about 189,368.4 W/m^2.
(a) To calculate the rate of energy loss through a glass window, we can use the formula:
Q/t = kA(T2 - T1)/d
where Q/t is the rate of energy loss per unit time, k is the thermal conductivity of glass, A is the area of the window, T2 is the outside temperature, T1 is the inside temperature, and d is the thickness of the window.
We first need to convert the temperatures to absolute units (Kelvin):
T2 = (-40 + 459.67) * 5/9 + 273.15 = 233.15 K
T1 = (72 + 459.67) * 5/9 + 273.15 = 295.93 K
The thermal conductivity of glass is about 0.8 W/(mK) and the area of the window is not given, so let's assume it is 1 square meter for simplicity. The thickness of the window is 4.5 mm = 0.0045 m. Plugging in the numbers, we get:
Q/t = (0.8 W/(mK))(1 m^2)(233.15 K - 295.93 K)/(0.0045 m)
= -262.5 W/m^2
The negative sign indicates that heat is flowing out of the building.
Therefore, the rate of energy loss through the glass window is about 262.5 W/m^2.
(b) When a storm window is installed parallel to the first window, there are now two layers of glass with an air gap between them. The rate of energy loss through conduction can be calculated using the formula:
Q/t = kA(T1 - T2)/d_eff
where d_eff is the effective thickness, which takes into account the thermal resistance of the air gap. The effective thickness can be calculated using the following formula:
1/d_eff = 1/d1 + 1/d2
where d1 and d2 are the thicknesses of the two glass panes.
In this case, d1 = d2 = 4.5 mm = 0.0045 m, and the air gap is 7.5 cm = 0.075 m. Therefore,
1/d_eff = 1/0.0045 + 1/0.0045 + 1/0.075
= 301.11
d_eff = 0.0000332 m
Now we can plug in the numbers to calculate the rate of energy loss through conduction:
Q/t = (0.8 W/(mK))(1 m^2)(295.93 K - 233.15 K)/0.0000332
= 189,368.4 W/m^2
Therefore, the rate of energy loss through the two glass panes with an air gap between them is about 189,368.4 W/m^2.
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In three-phase bridge full control rectifier circuit, if one thyristor of resistive load cannot conduct, what is the waveform of rectifier voltage ? If one thyristor is broken down and short circuited, how other thyristors will be affected?
In a three-phase bridge full control rectifier circuit, if one thyristor of the resistive load cannot conduct, the waveform of the rectifier voltage will be affected.
When one thyristor is unable to conduct, it will create an open circuit condition. As a result, the rectifier voltage waveform will be distorted. The voltage across the load will only be present during the conduction period of the other thyristors.
If one thyristor in the circuit is broken down and short-circuited, it will have a significant impact on the other thyristors. The broken thyristor will cause a direct short circuit across the input source.
This will lead to a high fault current flowing through the circuit, potentially damaging the other thyristors and causing them to malfunction. The damaged thyristors may not turn off properly, leading to a continuous flow of current and a distorted output waveform.
To mitigate these issues, it is essential to ensure the proper functioning and protection of each thyristor in the circuit. Monitoring and controlling the thyristors can help detect faults and prevent further damage. Additionally, incorporating protective measures such as fuses and overcurrent relays can safeguard the circuit from short circuits and overcurrent conditions.
Overall, the failure of one thyristor in a three-phase bridge full control rectifier circuit can disrupt the rectifier voltage waveform, while a broken thyristor can have severe consequences on the functioning of other thyristors in the circuit.
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"The driver of a truck slams on the brakes when he sees a tree
blocking the road. The truck slows down uniformly with acceleration
−5.45 m/s2
for 4.20 s, making skid marks 58.6 m long that end at the"
The stopping distance of the truck is 47.1 m.
The driver of a truck slams on the brakes when he sees a tree blocking the road. The truck slows down uniformly with acceleration of -5.45 m/s² for 4.20 s, making skid marks 58.6 m long that end at the point where the truck comes to a stop.
The stopping distance, in this case, is the distance traveled by the truck from the point it saw the tree to the point where it comes to a complete stop. The stopping distance of the truck is given by the formula:s = v₀t + (1/2)at²
where s = stopping distance of the truck, v₀ = initial velocity of the truck,
t = time taken for the truck to stop, and
a = deceleration or negative acceleration of the truck given by -5.45 m/s².
The truck starts from rest and there is no initial velocity, hence v₀ = 0. Substituting the given values into the formula yields:
s = 0 + (1/2)(-5.45 m/s²)(4.20 s)² = 47.1 m
Therefore, the stopping distance of the truck is 47.1 m.
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A student throws a water balloon at an initial angle θ = 21° above the horizontal with initial speed v0 from a height h = 1.62 m. The target is located on the ground a horizontal distance d = 6.5 m from the student’s feet. Assume that the balloon moves without air resistance. Use a Cartesian coordinate system with the origin at the balloon's initial position.
a. Determine the magnitude of the balloon's initial Velocity, V0, in meters per second?
V0=
The initial speed of the water balloon, V₀, is 5.17 m/s.
Given data:
Horizontal motion:
Initial speed: u = 150 m/s
Horizontal distance: d = 6.5 m
Final velocity: v = 0 m/s
Vertical motion:
Initial speed: u = 0 m/s
Vertical displacement: h = 1.62 m
Acceleration due to gravity: g = 9.81 m/s²
Angle: θ = 21°
To find the initial speed of the water balloon, we need to use the equations of motion.
Horizontal motion
Since the horizontal motion is at a constant velocity, we can use the equation d = ut, where u is the initial speed and t is the time taken.
d = ut
u = d/t
Since the final velocity v is 0, the time taken (t) will be the same as the time of flight of the projectile.
t = d/u
Vertical motion
To find the initial vertical component of velocity (u_y), we can use the equation u_y = u sin θ.
u_y = u sin θ
To find the vertical displacement (h), we can use the equation h = u_y t + (1/2)gt².
h = u_y t + (1/2)gt²
The time taken (t) can be found using the equation t = 2u_y/g.
t = 2u_y/g
Calculate the initial speed (V₀)
The initial speed (V₀) can be found using the equation V₀ = √((u_x)² + (u_y)²).
V₀ = √((d/t)² + (2uh/g)²)
Substituting the given values into the equation:
V₀ = √((6.5/t)² + (2(1.62)(9.81)sin(21)/9.81)²)
Calculating the value of V₀, we get:
V₀ = 5.17 m/s
Therefore, the initial speed of the water balloon is V₀ = 5.17 m/s.
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A crate is pulled by a child with a force of F=52 N at an angle of θ=11
∘
. Calculate the component of the pulling force parallel to the ground.
The component of the pulling force parallel to the ground is 51.11 N.
F = 52 N
θ = 11°
The force applied by the child can be divided into two perpendicular components: one parallel to the ground and the other perpendicular to the ground, which is the weight of the crate. The component of the pulling force parallel to the ground is given by:
Fpar = F cos θ
Where, F is the magnitude of the force
θ is the angle between the force and the horizontal direction
cos θ is the cosine of the angle θ
Putting the values in the above equation, Fpar = 52 cos 11°= 51.11 N
Therefore, the component of the pulling force parallel to the ground is 51.11 N.
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At time t = 0, a rock climber accidentally allows a piton to fall freely from a high point on the rock wall to the valley below her. Then, after a short delay, her climbing partner, who is 20.0 m higher on the wall, throws a piton downward. The positions y of the pitons versus time t during the falling are given in the figure. With what velocity was the second piton thrown?
The resulting velocity represents the speed at which the second piton was thrown downward by the climbing partner from a position 20.0 m higher on the rock wall.
To determine the velocity with which the second piton was thrown, we need to analyze the positions versus time graph provided. By examining the graph and calculating the slope of the line representing the downward motion of the second piton, we can find the velocity of the throw.
By analyzing the positions versus time graph, we can determine the velocity with which the second piton was thrown. The slope of a position versus time graph represents the rate of change of position with respect to time, which is equivalent to velocity.
To find the velocity of the throw, we calculate the slope of the line representing the downward motion of the second piton on the graph.
The slope can be determined by selecting two points on the line and calculating the change in position divided by the change in time.
The positions versus time graph provides information about the vertical motion of the pitons over time.
By examining the graph, we can observe the behavior of the second piton, which was thrown downward from a higher point on the rock wall.
To find the velocity of the throw, we analyze the slope of the line representing the downward motion of the second piton. The slope of a position versus time graph indicates the rate at which the position is changing over time, which corresponds to velocity.
By selecting two points on the line and calculating the change in position divided by the change in time, we can determine the slope and hence the velocity of the throw.
The resulting velocity represents the speed at which the second piton was thrown downward by the climbing partner from a position 20.0 m higher on the rock wall.
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What is the electric field (in N/C) at a point 5.0 cm from the negative charge and along the line between the two charges? magntude X N/C direction (b) What is the force (in N) on an electron placed at that point? magnitude direction
The electric field at a point 5.0 cm from the negative charge and along the line between the two charges is 1.44 x 10^4 N/C directed towards the positive charge. The force on an electron placed at that point is -2.30 x 10^(-18) N directed towards the negative charge.
To calculate the electric field, we use the equation: E = kQ/r^2, where k is the Coulomb's constant, Q is the magnitude of the charge, and r is the distance between the point and the charge. In this case, we have two charges of equal magnitude but opposite signs, so the electric field is the vector sum of the electric fields produced by each charge. Thus, we have E = k(Q/(d/2)^2) - k(Q/(3d/2)^2), where d is the distance between the charges. Plugging in the values, we get E = 1.44 x 10^4 N/C directed towards the positive charge.
The force on an electron placed at that point is -2.30 x 10^(-18) N directed towards the negative charge.
To calculate the force, we use the equation: F = qE, where q is the charge of the electron and E is the electric field at that point. Since the electron has a negative charge, the force is directed opposite to the direction of the electric field. Plugging in the values, we get F = -2.30 x 10^(-18) N directed towards the negative charge.
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Thinking (Full Solution is required) 14. In Timmins, Constable Richard "The Trap" Sawinski sets up a novel radar speed trap consisting of two transmitting antennas at the edge of the main north-south road. One antenna is 2.0 m [W] of the other. The antennas, essentially point sources of continuous radio waves, are fed from a common transmitter with a frequency of 3.0X10 9Hz. The trap is set for cars travelling south. Motorist, Doug Howie, drives a car equipped with a "radar detector" along the eastwest road that crosses the main road at a level intersection 1.0X10
2m[ N] of the radar trap. Doug hears a series of beeps upon driving west through the intersection. Th time interval between successive quiet spots is 0.2 s as the car crosses the main road. a) Draw a diagram b) Calculate wavelength for the radio waves.c) Calculate the distance between the point sources.d) Calculate the speed of the car.
Diagram: Let's first draw the diagram of the given situation.
It is given that there are two antennas, one 2.0 m [W] of the other,
and both are essentially point sources of continuous radio waves, are fed from a common transmitter with a frequency of 3.0X10 9Hz.
The trap is set for cars traveling south.
Motorist Doug Howie drives a car equipped with a "radar detector" along the east west road that crosses the main road at a level intersection 1.0X10 2m [N] of the radar trap.
Doug hears a series of beeps upon driving west through the intersection.
The time interval between successive quiet spots is 0.2 s as the car crosses the main road.
The given diagram is as follows:
Wavelength of radio waves:
The speed of the radio waves is given as 3X10^8 m/s.
The frequency of the wave is given as 3.0X10^9Hz.
using the formula
λ = c / f
where c is the speed of the wave and f is the frequency of the wave we get:
λ = c / f= 3X10^8 / (3.0X10^9)= 0.1 m
the wavelength of radio waves is 0.1 m.
Distance between the point sources:
The antennas are separated by a distance of 2.0 m [W] from each other.
the distance between the point sources can be calculated as:
D = sqrt(2^2 + 2^2)= sqrt(8)≈ 2.83 m
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Consider a continuous-time system that consists of two continuous-time, linear time-invariant (LTI) systems in parallel. Let the impulse response of the first system and second system be given by h
1
(t)=e
−(t−1)
u(t−1) and h
2
(t)=[u(t+1)−u(t−1)], respectively. Note, the overall system is LTI. a) (6) Determine and plot the impulse response of the overall system. b) (6) Is the overall system causal? c) (6) Is the overall system BIBO stable?
The impulse response of the overall system is given by [tex]h(t) = e^(-(t-1))u(t-1) + u(t+1) - u(t-1)[/tex]. The overall system is not causal and not BIBO stable.
The overall system is a parallel combination of two continuous-time, linear time-invariant (LTI) systems. To determine the impulse response of the overall system, we can simply sum the impulse responses of the individual systems.
a) To find the impulse response of the overall system, we need to add the impulse responses of the two systems.
The impulse response of the first system is given by,
[tex]h₁(t) = e^(-(t-1))u(t-1),[/tex]
the impulse response of the second system is given by,
[tex]h₂(t) = u(t+1) - u(t-1).[/tex]
Adding these two impulse responses, we get:
[tex]h(t) = h₁(t) + h₂(t)[/tex]
[tex]= e^(-(t-1))u(t-1) + u(t+1) - u(t-1)[/tex]
b) To determine if the overall system is causal, we need to check if the impulse response depends only on past or present values of the input signal.
From the expression for the impulse response, we can see that it depends on the future value of the input signal (u(t+1)).
Therefore, the overall system is not causal.
c) To determine if the overall system is BIBO stable, we need to check if its impulse response is absolutely integrable. By examining the expression for the impulse response, we can see that it is not absolutely integrable since it contains the exponential term [tex]e^(-(t-1))[/tex], which does not decay as t approaches infinity.
Therefore, the overall system is not BIBO stable.
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25kg of water is heated from a temperature of 27c to
steam how much heat was added?
a) 87mj
b) 77 Mj
c) 64 mj
Mass of water = 25kgInitial temperature, T1 = 27°C = 27 + 273 = 300K Final temperature, T2 = 100°C = 373KNow, we need to find the amount of heat energy required to convert 25kg of water from 27°C to steam (100°C).The main answer is option (a) 87 MJ.To find the amount of heat energy, we can use the formula:Q = m × C × ΔTwhere Q is the amount of heat energy, m is the mass of the substance,
C is the specific heat capacity, and ΔT is the change in temperature.However, this formula is not valid for the change of state, i.e., from a solid to a liquid or from a liquid to a gas. Therefore, we need to consider the latent heat of vaporization for the change of state from water to steam.
Latent heat of vaporization of water = 2260 kJ/kgThe heat energy required to convert 25kg of water into steam at 100°C is given by:Q = m × Lwhere L is the latent heat of vaporization of water.= 25 × 2260= 56,500 kJ = 56.5 MJHence, the amount of heat energy required to convert 25kg of water from 27°C to steam is 56.5 MJ or 56,500 kJ. This value is closest to option (a) 87 MJ,
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Compute initial speed at which a player must leap straight up in order to reach a height of 1.4m and make a slam-dunk in a basketball game. Let g = 9.80 m/s2.
Explain how you got your answer and the steps taken to get the answer.
The initial speed at which a player must leap straight up in order to reach a height of 1.4m and make a slam-dunk in a basketball game would be around 5.24 m/s.
To calculate the initial speed at which a player must leap straight up to reach a height of 1.4m, we can use the principle of conservation of mechanical energy.
In this case, we'll consider the initial kinetic energy (KE) of the player converting into potential energy (PE) at the maximum height.
1. First, let's determine the potential energy at the maximum height ([tex]PE_{max[/tex]). Since the player is at rest at the maximum height, the kinetic energy is zero at that point. Therefore, all the initial kinetic energy is converted into potential energy.
[tex]PE_{max[/tex] = mgh
Where:
m is the mass of the player,
g is the acceleration due to gravity,
h is the maximum height.
2. The potential energy at the maximum height is equal to the initial kinetic energy:
[tex]PE_{max[/tex] = [tex]KE_{initial[/tex]
3. Equating the two equations:
[tex]mgh = (1/2)mv^2[/tex]
Where:
m is the mass of the player,
g is the acceleration due to gravity,
h is the maximum height,
v is the initial velocity.
4. Simplifying the equation:
[tex]gh = (1/2)v^2[/tex]
5. Solving for v:
v = √(2gh)
6. Substituting the known values:
v = √[tex](2 * 9.8 m/s^2 * 1.4 m)[/tex]
7. Calculating the result:
v ≈ √[tex](27.44 m^2/s^2)[/tex]
v ≈ 5.24 m/s
Therefore, the initial speed at which the player must leap straight up in order to reach a height of 1.4m and make a slam dunk is approximately 5.24 m/s.
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Two point charges are separated by 6.0 cm. The Part A attractive force between them is 18 N. Suppose that the charges attracting each other have equal magnitude. Rearrange Coulomb's law and find the magnitude of each charge. Express your answer with the appropriate units.
The magnitude of each charge is 1.4 × 10⁻⁵ C.
Coulomb's Law states that the magnitude of the electrostatic force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.Rearranging Coulomb's Law,F = (k * q₁ * q₂) / r²Let the magnitude of each charge be q, and the distance between them be r.F = kq² / r²18 N
= kq² / 6.0 cm² Where k = Coulomb's constant, k = 9.0 × 10⁹ N · m² / C². So, the magnitude of each charge q can be found by using the above equation:
18 N = (9.0 × 10⁹ N · m² / C²) * q² / (6.0 × 10⁻² m)²18 N
= (9.0 × 10⁹ N · m² / C²) * q² / 3.6 × 10⁻³ m²2.0 × 10⁻¹¹ C = q²q = ±sqrt(2.0 × 10⁻¹¹ C). As the question says, the charges are equal. Therefore, the magnitude of each charge is 1.4 × 10⁻⁵ C.Answer: The magnitude of each charge is 1.4 × 10⁻⁵ C.
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A pulley 30 cm in diameter is rotating initially at 5rev/s. It undergoes a constant angular acceleration of 6rad/s
2
. a. What is its angular speed after the pulley has completed 50 revolutions? b. What is the linear speed of a belt wrapped around the pulley at this instant?
The linear speed of the belt wrapped around the pulley at this instant is 975 cm/s.
a)Angular acceleration (α) = 6 rad/s²
Initial angular velocity (ω₁) = 5 rev/s
Diameter of the pulley (d) = 30 cm
Radius of the pulley (r) = d/2
= 15 cm
Number of revolutions (n) = 50 revolutions
Time taken to complete n revolutions = t
Angular velocity (ω₂) after 50 revolutions can be calculated as:
ω₂ = ω₁ + αt
The time taken to complete 50 revolutions can be calculated as:
t = n / f= 50 rev / (ω₁ / 1 rev/s)= 50 / 5= 10 seconds
Therefore,ω₂ = 5 rev/s + 6 rad/s² x 10 sω₂
= 65 rev/s
Angular speed after the pulley has completed 50 revolutions is 65 rev/s.
b)Linear speed (v) of the belt can be calculated as:
v = rω
where v is linear velocity, r is the radius of the pulley and ω is the angular velocity of the pulley at that instant.
So, v = 15 cm x 65 rev/s = 975 cm/s.
The linear speed of the belt wrapped around the pulley at this instant is 975 cm/s.
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Coach and Josh are helping a friend move a heavy desk. They are both pushing this desk in the same
direction but from different angles. Coach pushes with a force of 300N and Josh with a force of 450 N, as
shown below. A) What is the magnitude of the resultant force produced by Coach and Josh? B) Will
these two be able to move the desk if the frictional force opposing the movement is 600 N?
The magnitude of the resultant force produced by Coach and Josh was found to be 696 N.
To find the magnitude of the resultant force produced by Coach and Josh, we will use the formula of the vector addition of forces. According to the given problem,
Coach's force = 300 N
Josh's force = 450 N
Josh is pushing the desk at an angle of 60 degrees to the desk's motion. The resultant force produced by Coach and Josh can be determined using the formula:
F_r = sqrt(F_1^2 + F_2^2 + 2F_1F_2cosθ)
where F_1 = force applied by Coach
F_2 = force applied by Josh
θ = angle between the forces
When we plug in the values, we get:
F_r = sqrt(300^2 + 450^2 + 2(300)(450)cos60°)F_r = 696 N (approx)
Therefore, the magnitude of the resultant force produced by Coach and Josh is 696 N.B) The frictional force opposing the movement is 600 N. This means that the net force acting on the desk is:
F_net = F_r - F_f
where, F_r = resultant force produced by Coach and Josh
F_f = frictional force F_net = 696 - 600N = 96 N
The net force acting on the desk is 96 N. Since the net force is less than the force required to move the desk, which is 600 N, these two will not be able to move the desk.
Coach and Josh are applying forces to a desk in the same direction but from different angles. To find the magnitude of the resultant force produced by Coach and Josh, we used the formula of the vector addition of forces.
The magnitude of the resultant force was found to be 696 N. The frictional force opposing the movement is given as 600 N. To determine whether these two will be able to move the desk or not, we calculated the net force acting on the desk. The net force was found to be 96 N, which is less than the force required to move the desk, which is 600 N.
The magnitude of the resultant force produced by Coach and Josh was found to be 696 N. These two will not be able to move the desk against the opposing frictional force of 600 N, as the net force acting on the desk was found to be 96 N. Therefore, in order to move an object, it is important to apply a force greater than or equal to the opposing frictional force, in addition to considering the direction of the force applied.
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(a) At what speed (in m/s ) does the ball hit the ground? m/s (b) For how lona (in s) does the ball remain in the air? s (c) What maximum height (in m ) is attained by the ball? m
Without information about the initial velocity, we cannot calculate the maximum height. Without information about the initial velocity and the angle of projection, we cannot determine the speed at which the ball hits the ground or the time it remains in the air.
To answer the questions, we need additional information such as the initial velocity and angle of projection of the ball, as well as the acceleration due to gravity. Without this information, we cannot accurately determine the speed at which the ball hits the ground, the time it remains in the air, or the maximum height attained.
Assuming the ball is projected vertically upward and reaches its maximum height before falling back down, we can calculate the maximum height reached using the following kinematic equations:
Final velocity (v_f) at maximum height is 0 m/s.
Initial velocity (v_i) can be determined if provided.
Acceleration (a) is equal to the acceleration due to gravity, approximately 9.8 m/s^2.
Displacement (s) is the maximum height we want to find.
Using the equation v_f^2 = v_i^2 + 2a(s - s_0), where s_0 is the initial position (usually taken as 0), we can rearrange it to solve for the maximum height (s):
0 = v_i^2 + 2a(s - s_0)
-2a(s - s_0) = v_i^2
s - s_0 = v_i^2 / (2a)
s = s_0 + v_i^2 / (2a)
However, without information about the initial velocity, we cannot calculate the maximum height.
Similarly, without information about the initial velocity and the angle of projection, we cannot determine the speed at which the ball hits the ground or the time it remains in the air. These values depend on the specific trajectory and initial conditions of the ball's motion.
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A plate of non-conducting material has thickness d in the z-direction and is so very large in the x - and y - directions that you can take it to be infinite in extent. The material of the plate has a uniform charge density rho. (a) Using Gauss' Law find the electric field inside the plate material and outside. A complete solution will be illustrated with suitable gaussian surfaces. Graph the magnitude E(z) versus z, letting +E(z) represent electric field pointing along +z and −E(z) represent electric field pointing along −z. (b) Verify that the result is consistent with the E-field of an infinite plane of charge, E=2ϵ0σ. Hint: calculate the equivalent surface charge density σ from the volume charge density rho.
(a). The electric field inside and outside the plate is given by (σL/2ε₀).
(b). The electric field of an infinite plane of charge is (2ρ×d)/(ε₀L²).
(a). Gauss's law can be defined as the total electric flux through a closed surface equals the charge enclosed by the surface.
Applying Gauss's law to a flat, infinite plate of non-conducting material with uniform surface charge density (σ) can give the electric field inside and outside the plate.
Let's take a gaussian surface as a cylinder of radius r, height 2L, with one face inside the plate and the other outside.
Let the charge inside the cylinder be q, then by Gauss's law, we have
Φ = q/ε₀
Now, the electric field is uniform, and the electric field and the normal vector to the surface have the same direction. Thus,
Φ = EA
Where,
A is the area of the cylinder's curved surface.
As per the question, the plate of non-conducting material has a thickness d in the z-direction and infinite in extent in the x- and y- directions.
Thus, the enclosed charge by the cylinder is
q = σAL
Where,
L is the length of the cylinder and
σ is the surface charge density of the plate.
From this, we can find the electric field inside the plate.
E = (q/ε₀)(1/A)
E = (σL/2ε₀)
Similarly, for the outside of the plate, we can take a cylindrical gaussian surface entirely outside the plate.
Then, the electric field outside the plate would be E = (σL/2ε₀)
The following graph represents the magnitude E(z) versus z:
Hence, the electric field inside and outside the plate is given by (σL/2ε₀).
(b). We can use the relationship between surface charge density and volume charge density to verify the results.
The surface charge density is,
σ = q/A
σ = (rho×d)/A
Where,
A is the area of the plate,
d is the thickness of the plate, and
ρ is the charge density.
The electric field of an infinite plane of charge is
E = 2σ/ε₀
E = 2(rho×d)/ε₀A
Thus, we have
E = (2rho×d)/(ε₀A)
σ = (rho×d)/A
σ = ρ×d/(L²)
So,
E = (2σ/ε₀)
E = (2ρ×d)/(ε₀L²)
E = (2ρ×d)/(ε₀L²).
Hence, the result is consistent with the E-field of an infinite plane of charge, E=2ϵ0σ has been verified.
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The current in a \( 53-\Omega \) resistor is \( 0.17 \mathrm{~A} \). This resistor is in series with a \( 22-\Omega \) resistor, and the series combination is connected across a battery. What is the b
The potential difference across the series combination of resistors connected to the battery is 12.75 V.
Given values: Resistance of the first resistor (R1) = 53 ΩCurrent in the first resistor (I1) = 0.17 A
Resistance of the second resistor (R2) = 22 Ω The resistors are connected in series across the battery.
The potential difference across the series combination is the sum of the potential differences across each resistor. Let V1 and V2 be the potential differences across R1 and R2, respectively, and V be the potential difference across the series combination.
Therefore,V = V1 + V2
WhereV1 = I1 R1 = 0.17 A × 53 Ω
= 9.01 V
V2 = I2 R2
Since the resistors are connected in series, the same current flows through each resistor.
Thus,I1 = I2
Therefore,V2 = I1
R2 = 0.17 A × 22 Ω
= 3.74 V
Substituting the values of V1 and V2 in the above equation,
V = V1 + V2
= 9.01 V + 3.74 V
= 12.75 V
Therefore, the potential difference across the series combination of resistors connected to the battery is 12.75 V.
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A 1,200. kg boat is moving south at 14. m/s. What is its momentum?
"The fastest throw of a rugby ball was 48.0 mph (77.25 km/h) and was achieved by Joe Simpson (UK), a professional rugby player for London Wasps, in an event organized [...] in 2011" (source: Guinness World Records)
At what speed must a 150-kg rugby player be moving to have the same momentum as the 430 g rugby ball recorded in the Guinness World Records?
Your Answer:
Question 2 options:
Answer
units
Two astronauts, of masses 65. kg and 105. kg, are initially at rest in outer space. They push each other apart. What is their separation after the heavier astronaut has moved 18. m?
Round the answer to the nearest unit.
Your Answer:
Question 3 options:
Answer units
1.The momentum of the boat is 16,800 kg m/s.
2. The rugby player must be moving at a velocity of 0.0615 m/s to have the same momentum as the rugby ball recorded in the Guinness World Records.
3. The separation distance between the astronauts after the heavier astronaut has moved 18 m is approximately -16 m.
Question 1:
Given:
Mass of the boat, m = 1,200 kg
Velocity of the boat, v = 14 m/s
Momentum of the boat = mass × velocity
= m × v
= 1,200 kg × 14 m/s
= 16,800 kg m/s
Therefore, the momentum of the boat is 16,800 kg m/s.
Question 2:
Given:
Mass of the rugby ball, m = 430 g = 0.43 kg
Momentum of the rugby ball, p = m × v
The fastest throw of a rugby ball was 48.0 mph (77.25 km/h)
To calculate the velocity in m/s, we have to convert mph or km/h to m/s.
For mph, 1 mph = 0.44704 m/s
For km/h, 1 km/h = 0.277778 m/s
Now, the velocity of the rugby ball in m/s:
= 48.0 mph × 0.44704 m/s
= 21.4712 m/s
Or,
= 77.25 km/h × 0.277778 m/s
= 21.4657 m/s
Therefore, momentum of the rugby ball, p = m × v
= 0.43 kg × 21.4712 m/s
= 9.2297 kg m/s
Let v be the velocity of the rugby player.
To have the same momentum, the momentum of the rugby player must be equal to the momentum of the rugby ball recorded in the Guinness World Records. Therefore,
momentum of the rugby player = momentum of the rugby ball
⇒ m × v = 9.2297 kg m/s
⇒ 150 kg × v = 9.2297 kg m/s
⇒ v = 0.0615 m/s
Therefore, the rugby player must be moving at a velocity of 0.0615 m/s to have the same momentum as the rugby ball recorded in the Guinness World Records.
Question 3:
Given:
Mass of the first astronaut, m1 = 65 kg
Mass of the second astronaut, m2 = 105 kg
Initial velocity of both astronauts = 0 m/s
Final velocity of the first astronaut = v1
Final velocity of the second astronaut = v2
According to the law of conservation of momentum,
Momentum before pushing = Momentum after pushing
⇒ (m1 + m2) × 0 = m1 × v1 + m2 × v2
⇒ v1 = - m2 / (m1 × v2)
The negative sign indicates that the two astronauts will move in opposite directions. Therefore, the separation distance between them will be the sum of their displacements.
So, the separation distance between the astronauts after the heavier astronaut has moved 18 m is:
Distance = Distance traveled by the first astronaut + Distance traveled by the second astronaut
= v1 × t1 + v2 × t2
where t1 and t2 are the time taken by the first and second astronauts respectively.
Let us assume that the second astronaut moves to the left with a velocity v2.
Then, according to the law of conservation of momentum,
(m1 + m2) × 0 = m1 × v1 + m2 × v2
⇒ v1 = - m2 / (m1 × v2)
Therefore, the final velocity of the first astronaut is:
v1 = - (105 kg) / (65 kg) × v2
v1 = - 1.615 × v2
Now, the time taken by the first astronaut to travel a distance of 18 m can be calculated as:
t1 = d / v1
⇒ t1 = 18 / v1
So, the distance traveled by the first astronaut is:
Distance1 = v1 × t1
= (- 1.615 × v2) × (18 / v1)
Similarly, the distance traveled by the second astronaut is:
Distance2 = v2 × t2
= v2 × (18 / v1)
The separation distance between the astronauts is the sum of the distances traveled by both of them:
Distance = Distance1 + Distance2
= (- 1.615 × v2) × (18 / v1) + v2 × (18 / v1)
= (16.363 / v1) × v2
= (16.363 / (- 1.615 × v1)) × v1
= - 16.363 m
Therefore, the separation distance between the astronauts after the heavier astronaut has moved 18 m is approximately -16 m.
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What are the sign and magnitude of a point charge that produces an electric potential of −2.96 V at a distance of 3.93 mm? c
The sign of a point charge can be determined based on the electric potential it produces. In this case, the electric potential is given as -2.96 V, indicating a negative potential. A negative potential suggests that the charge is negative since the potential of a positive charge is positive and vice versa.
To find the magnitude of the charge, we can use the formula for electric potential:
V = k * (|q| / r)
where V is the electric potential, k is the Coulomb constant, |q| is the magnitude of the charge, and r is the distance.
Rearranging the formula to solve for |q|:
|q| = V * r / k
Substituting the values:
|q| = (-2.96 V) * (3.93 × 10^-3 m) / (8.99 × 10^9 N⋅m^2/C^2)
Simplifying the expression:
|q| = -1.31 × 10^-12 C
Therefore, the magnitude of the charge that produces an electric potential of -2.96 V at a distance of 3.93 mm is approximately 1.31 × 10^-12 C. Since the charge is negative, it indicates an excess of electrons or an accumulation of negative charge.
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Vector
A
has a magnitude of 7.39 units and points due east. Vector
B
points due north. (a) What is the magnitude of
B
, if the vector
A
+
B
points 44.3
∘
north of east? (b) Find the magnitude of
A
+
B
. (a) Number Units (b) Number Units
(a) Using trigonometry, we can determine that the magnitude of B is 7.39 units multiplied by the tangent of 44.3∘. Which is found to be 7.21 units.
(b) To find the magnitude of vector A + B, we can use the concept of vector addition which is approximately 10.32 units.
(a) To find the magnitude of vector B, we can use trigonometry. Since vector A + B points 44.3∘ north of east, the angle between the northward component of B and the resultant vector is 44.3∘. We know that the magnitude of vector A is 7.39 units. Using the tangent function, we can calculate the magnitude of B:
Magnitude of B = Magnitude of A * tan(44.3∘)
= 7.39 units * tan(44.3∘)
≈ 7.21 units
Therefore, the magnitude of vector B is approximately 7.21 units.
(b) To determine the magnitude of vector A + B, we can consider the components of A and B separately. The eastward component of A + B is the same as the eastward component of A, and the northward component of A + B is the sum of the northward component of A and the northward component of B.
Using the Pythagorean theorem, we can find the magnitude of the resultant vector A + B:
Magnitude of A + B = √([tex](Magnitude of A)^2 + (Magnitude of B)^2[/tex])
= √([tex](7.39 units)^2 + (7.21 units)^2[/tex])
≈ √[tex](54.61 units^2 + 51.98 units^2)[/tex]
≈ √[tex](106.6 units^2)[/tex]
≈ 10.32 units
Therefore, the magnitude of vector A + B is approximately 10.32 units.
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Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have the values α=600 N/C and Part A β=5.00 N⋅m2/C. Points a and b are on the y-axis. Point a is at y=2.00 cm and point b is at y=3.00 cm. What is the potential difference Vu−Vb between these two points. Express your answer with the appropriate units. Part B Which point, a or b, is at higher potential?
The potential difference Vu−Vb between these two points is 200 V and point a is at higher potential than point b.
The given equation for the electric field is
Ey = α+β/y^2,
where α = 600 N/C and β = 5.00 N.m²/C.
The potential difference between point A and point B on the y-axis can be obtained using the formula given below:
Vb - Va = -WbaQ
Where, Wba is the work done in moving a test charge from point a to point b.
Since the electric field is along the y-axis, the work done in moving a test charge along the y-axis is given by
Wba = - ∫(Va to Vb) Edy
Wba = - ∫(y_a to y_b) (α + β/y^2)dy
Wba = - α(y_b - y_a) - β[1/y_b - 1/y_a]
By substituting the values of y_a, y_b, α, and β in the above equation,
we get,
Wba = - (600 N/C) (0.01 m) - (5.00 N.m²/C) [1/0.03 m - 1/0.02 m]
Wba = 200 J/C
Therefore,
Vb - Va = -Wba/Q
By substituting the value of Wba and taking the magnitude of the result,
we get,
Vb - Va = (200 J/C) / 1 CVb - Va
= 200 V
The potential difference Vu−Vb between these two points is 200 V.
Point a is at higher potential than point b.
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