Answer:
28343.3 kgm/s
177145.625 N
Explanation:
u = Initial velocity = 19 m/s
v = Final velocity = -3.3 m/s (opposite direction)
m = Mass of car = 1271 kg
t = Time taken = 0.16 s
Impulse is given by
[tex]J=m(v-u)\\\Rightarrow J=1271(-3.3-19)\\\Rightarrow J=-28343.3\ kgm/s[/tex]
The magnitude of the impulse due to the collision is 28343.3 kgm/s
Force given by
[tex]F=\dfrac{J}{t}\\\Rightarrow F=\dfrac{-28343.3}{0.16}\\\Rightarrow F=-177145.625\ N[/tex]
The magnitude of the average force exerted on the automobile is 177145.625 N
Final answer:
The magnitude of the impulse during the collision is 28345.3 kg·m/s, and the magnitude of the average force exerted on the automobile is 177158 N.
Explanation:
In the crash test scenario described, the impulse experienced by the automobile can be determined by using the change in momentum due to the collision. Momentum (p) is the product of mass (m) and velocity (v), and impulse is the change in momentum which can be calculated using the initial (vi) and final velocities (vf) of the car:
Impulse = Change in momentum
= m × (vf - vi)
For the automobile:
Mass (m) = 1271 kgInitial velocity (vi) = 19 m/s (positive in the direction of motion)Final velocity (vf) = -3.3 m/s (negative since the car bounces back)Impulse = 1271 kg × (-3.3 m/s - 19 m/s)
= 1271 kg × (-22.3 m/s)
= -28345.3 kg·m/s (to one decimal placing)
The magnitude of the impulse is the absolute value:
28345.3 kg·m/s.
To calculate the average force exerted on the automobile, we use the fact that impulse also equals the average force (Favg) multiplied by the time of impact (t):
Impulse = Favg × t
Thus, the average force can be found by dividing the impulse by the collision time.
Favg = Impulse / t
= 28345.3 kg·m/s / 0.16 s
= 177158.1 N (to one decimal placing)
The magnitude of the average force is the absolute value: 177158 N.
If two planets orbit a star, but planet B is twice as far from the star as planet A, planet A will receive ____ times the flux that planet B receives.
Answer:
The nearest plant (A) receives 4 times more radiation from the farthest plant
Explanation:
The energy emitted by the star is distributed on the surface of a sphere, whereby intensity received is the power emitted between the area of the sphere
I = P / A
P = I A
The area of the sphere is
A = 4π r²
Since the amount of radiation emitted by the star is constant, we can write this expression for the position of the two planets
P = I₁ A₁ = I₂ A₂
I₁ / I₂ = A₂ / A₁
Suppose index 1 corresponds to the nearest planet,
r2 = 2 r₁
I₁ / I₂ = r₁² / r₂²
I₁ / I₂ = r₁² / (2r₁)²
I₁ / I₂ = ¼
4 I₁ = I₂
The nearest plant (A) receives 4 times more radiation from the farthest plant
If someone is riding in the back of a pickup truck and throws a softball straight backward, is it possible for the ball to fall straight down as viewed by a person standing at the side of the road? Under what condition would this occur? How would the motion of the ball appear to the person who threw it?
Yes, it is possible. The ball would fall straight down from the perspective of a person standing at the side of the road. The person who threw the ball would see it initially move backward and then fall vertically downward.
Explanation:Yes, it is possible for the ball to fall straight down as viewed by a person standing at the side of the road while someone is riding in the back of a pickup truck and throws a softball straight backward. This would occur under the condition that the ball is thrown with the same initial velocity as the truck's speed. When the ball is thrown straight backward with the same initial speed as the truck's, it will continue to move with the same speed in the backward direction relative to the truck. From the perspective of the person who threw the ball, they would see the ball initially move straight backward and then fall vertically downward due to the force of gravity.
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The ball will fall straight down as viewed from the side of the road if the velocities cancel each other out.
To the person who threw the ball, it would appear to travel straight backward relative to them.
If someone is riding in the back of a pickup truck and throws a softball straight backward, it is possible for the ball to fall straight down as viewed by a person standing at the side of the road. This will occur if the velocity at which the ball is thrown backward is equal to the velocity of the truck moving forward. In this case, the forward motion of the truck and the backward motion of the ball will cancel each other out.
For the person who threw the ball, the motion would still appear as if the ball was thrown straight backward relative to their frame of reference, assuming the truck is moving with constant velocity. The ball would move backward with the same speed at which it was thrown.
To summarize:
The ball will fall straight down as viewed from the side of the road if the velocities cancel each other out.
To the person who threw the ball, it would appear to travel straight backward relative to them.
A car initially going 61 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 7 seconds.
a) Graph the velocity from t = 0 to t = 7.
(b) How far does the car travel?
(c) How far does the car travel if its initial velocity is doubled, but it brakes at the same constant rate?
Answer:
See the attachment below for the graphics in part (a)
The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.
This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.
Explanation:
The full solution can be found in the attachment below.
Thank you for reading. I hope this post is helpful to you.
A 65.0 kg ice skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. What is the velocity of the skater?
a. -0.07 m/s
b. 0.15 m/s
c. 0.30 m/s
d. 0.07 m/s
Answer:
(d) 0.07 m/s
Explanation:
Given Data
Snowball mass m₁=0.15 kg
Ice skater mass m₂=65.0 kg
Snowball velocity v₁=32.0 m/s
To find
Velocity of Skater v₂=?
Solution
From law of conservation of momentum
[tex]m_{1}v_{1}=m_{2}v_{2}\\ v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s[/tex]
So Option d is correct one
The velocity of skater is [tex]0.07m/s[/tex]
Option d is correct.
Law of conservation of momentum:For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied.
The expression is given as,[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]
Given that, mass of ice skater [tex]m_{1}=65kg[/tex], mass of snow ball [tex]m_{2}=0.15kg[/tex] Velocity of snow ball [tex]v_{2}=32m/s[/tex]Substitute values in above expression.[tex]65*v_{1}=0.15*32\\\\v_{1}=\frac{0.15*32}{65} =0.07m/s[/tex]
The velocity of skater is [tex]0.07m/s[/tex]
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A Ferris wheel rotating at 20 rad/s slows down with a constant angular acceleration of magnitude 5.0 rad/s2. How many revolutions does it make while slowing down before coming to rest
Answer:
It takes 6.37 revolutions to stop.
Explanation:
The constant angular acceleration is negative if we choose the wheel direction of rotation as positive direction, so radial acceleration is α=5.0[tex] \frac{rad}{s^{2}}[/tex].Because the wheel is changing its velocity and we already know radial acceleration we should use the Galileo's kinematic rotational equation:
[tex]\omega^{2}=\omega_{i}^{2}+2\alpha\varDelta\theta [/tex]
with [tex] \omega[/tex] the final angular velocity (is zero because the wheel comes to rest), [tex] \omega_{i}[/tex] the initial angular velocity and Δθ the angular displacement. Solving (1) for Δθ :
[tex]\varDelta\theta=\frac{\omega^{2}-\omega_{i}^{2}}{2\alpha} [/tex]
[tex]\varDelta\theta=\frac{0-20^{2}}{(2)(-5.0)}=40rad [/tex]
The angular displacement can be converted to revolution knowing that 1 revolution is 2π rad:
[tex]40rad=\frac{40rad}{2\pi\frac{rad}{rev}} [/tex]
[tex] 40rad=6.37 rev[/tex]
Number of revolution take place is 6.4 revolution
Given that;Velocity of rotating wheel = 20 rad/s
Acceleration of magnitude = 5.0 rad/s²
Find:Number of revolution take place
Computation:Using Third equation of motion;
v² - u² = 2as20² - 0² = 2(5)(s)
400 = (10)(s)
Total distance = 40 rad
Number of revolution take place = Total distance / 2π
Number of revolution take place = 40 / 2(3.14)
Number of revolution take place = 40 / 6.28
Number of revolution take place = 6.369
Number of revolution take place = 6.4 revolution
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By what distance do two objects carrying 1.0 C of charge each have to be separated before the electric force exerted on each object is 3.0 N ?
To solve this problem we will apply the definition of electrostatic force. From the variables present there and explained later we will find the value of the distance reorganizing said expression, that is
[tex]F = \frac{k q_1 q_2}{d^2}[/tex]
Here
k = Coulomb's constant
[tex]q_{1,2}[/tex] = Charge of each object
d = Distance
Replacing our values we have that
[tex]3 = \frac{(9*10^9)(1)(1)}{d^2}[/tex]
Rearranging and solving for the distance we have
[tex]d = 54772.25m[/tex]
Therefore the distance between the two objects is 54772.25m
Two charged small spheres are a distance R apart and exert an electrostatic force F on each other. If the distance is halved to R/5 , the force exerted on each sphere will be..
a) 25F
b) 5F
c) F/5
d) F/25
Answer:
a) 25F
Explanation:
Assuming that the two small spheres can be modeled as point charges, according to Coulomb's law, the magnitude of the electrostatic force is given by:
[tex]F=\frac{kq_1q_2}{R^2}[/tex]
In this case, we have [tex]R'=\frac{R}{5}[/tex]:
[tex]F'=\frac{kq_1q_2}{R'^2}\\F'=\frac{kq_1q_2}{(\frac{R}{5})^2}\\F'=25\frac{kq_1q_2}{R^2}\\F'=25F[/tex]
a ferry which runs at 12m/s in still water travels between towns a and b on a river which flows south at 9 m/s. the ferry leavs a and heads due east but reaches the east bank at C. it then heads north to B. whatis the velocity of the ferry as it traverses route ac?
Answer:
= 15 m/s
Explanation:
Considering right side(west) as positive x-axis and south as negative y-axis.
velocity of boat in still water [tex]v_b=12\hat{i}[/tex]
velocity of stream [tex]v_s=-9\hat{j}[/tex]
now relative velocity of boat w.r.t. stream [tex]v_{b/s}=12\hat{i}+9\hat{j}[/tex]
this velocity with which ac distance will be covered.
therefore magnitude of [tex]v_{b/s} =\sqrt{12^2+9^2}[/tex]
= 15 m/s
A 8-hp (shaft) pump is used to raise water to an elevation of 15 m. If the mechanical efficiency of the pump is 82 percent, determine the maximum volume flow rate of water.
The maximum Volume Flow Rate of water the pump can provide, given an efficiency of 82% and an elevation of 15 m, is approximately 0.033 L/s.
First, we must convert the pump's horsepower to a more usable unit in this context - like watts. In physics, 1 horsepower equals roughly 746 watts. Therefore, the pump has power of 8*746 = 5968 watts.
Given the mechanical efficiency (ME) and the height (h), the maximum work the pump can do is given by M.E. * Power. So, the pump does work of 0.82*5968 = 4895.76 watts.
The work done on the water by the pump is equal to the change in potential energy of the water, PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (15 m). With rearranging, you could express m = Power/(g * h). But we're looking for the volume flow rate, not the mass flow rate, so we need to convert mass to volume. Since the density of water (ρ) is 1 kg/L, the volume flow rate = m/ρ = Power/(g * h * ρ).
Substituting all known values, we get: Volume flow rate = 4895.76 W /(9.8 m/s^2 * 15 m * 1 kg/L) = 0.033 L/s.
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A sinusoidal sound wave moves through a medium and is described by the displacement wave function
s(x, t) = 2.00 cos (15.7x ? 858t)
where s is in micrometers, x is in meters, and t is in seconds. Find (a) the amplitude, (b) the wavelength, and (c) the speed of this wave.
Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x
Answer
given,
displacement wave function
s(x, t) = 2.00 cos (15.7 x + 858 t)
now, comparing the wave equation with general equation.
s(x, t) = A cos (k x + ω t)
where A is the amplitude of the wave in micrometer.
now,
a) Amplitude of the wave
A = 2 x 10⁻⁶ μ m
b) [tex]\lambda = \dfrac{2\pi}{k}[/tex]
k = 15.7 m
[tex]\lambda = \dfrac{2\pi}{15.7}[/tex]
λ = 0.4 m
c) wave speed
[tex]v = \dfrac{\omega}{k}[/tex]
[tex]v = \dfrac{858}{15.7}[/tex]
v = 54.65 m/s
d) For instantaneous displacement
Assuming the position and time is given as
x = 0.05 m and t = 3 m s
now,
s(x, t) = 2.00 cos (15.7 x 0.05 + 858 x 3 x 10⁻³ )
s(x, t) = 2.00 cos (3.359)
s(x,t) = -1.95 μ m
The amplitude of the wave is 2.00 micrometers, the wavelength is 15.7, and the speed of the wave is approximately 54.65 m/s.
Explanation:The displacement wave function is given as s(x, t) = 2.00 cos (15.7x - 858t). In this wave function, the amplitude is the coefficient in front of the cosine function, which is 2.00 micrometers. The wavelength can be determined by finding the coefficient of the x term inside the cosine function, which is 15.7. The speed of the wave can be calculated by dividing the angular frequency (in this case, 858) by the wave number (in this case, 15.7), resulting in a speed of approximately 54.65 m/s.
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An oscillator creates periodic waves on a stretched string.
If the period of the oscillator doubles, what happens to the wavelength and wave speed?
a. The wavelength doubles but the wave speed is unchanged.
b. The wavelength is halved but the wave speed is unchanged.
c. The wavelength is unchanged but the wave speed doubles.
Answer:
A. The wavelength doubles but the wave speed is unchanged
Explanation:
The relationship between the period and wavelength is direct. Doubling the period of the oscillator will correspondingly double the wavelength but the wave speed is unaffected
An oscillator creates periodic waves on a stretched string. If the period of the oscillator doubles, then the wavelength doubles but the wave speed is unchanged. So option A is correct here.
When the period of the oscillator doubles, it means that the time it takes for one complete oscillation or cycle of the wave doubles. The period of a wave is inversely proportional to its frequency. If the period doubles, the frequency is halved. The wavelength of a wave is the distance between two consecutive crests or troughs. The wavelength of a wave is inversely proportional to its frequency. When the frequency is halved, the wavelength doubles to maintain the relationship.
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If the accelerating voltage va were 250 volts, what would be the speed of the electrons emerging from the gun?
Answer:
[tex]v=9.34*10^{6}\frac{m}{s}[/tex]
Explanation:
According to the law of conservation of energy, the amount of energy an electron gains after being accelerated is equal to its kinetic energy, that is, the electrical potential energy is converted into kinetic energy :
[tex]U=K\\eV=\frac{mv^2}{2}[/tex]
Solving for v and replacing the given voltage:
[tex]v=\sqrt\frac{2eV}{m}\\v=\sqrt\frac{2(1.6*10^{-19}C)250V}{9.1*10^{-31}kg}\\v=9.34*10^{6}\frac{m}{s}[/tex]
A small object is attached to the end of a relaxed, horizontal spring whose opposite end is fixed. The spring rests on a frictionless surface. Let the initial position of the object be defined as x-0. The object is pulled to position x = A and then released, after which it undergoes simple harmonic motion.
In one full cycle of its motion, the total distance traveled by the object is:
1) A
2) A/2
3) 2A
4) 4A
5) A/4
Answer:4A
Explanation:
Given
Mass is displace x= A units from its mean position x=0'
When it is set to free it will oscillate about its mean position with maximum amplitude A i.e. from x=-A to x=A
One cycle is completed when block returns to its original position
so first block will go equilibrium position x=0 and then to x=-A
from x=-A it again moves back to x=0 and finally back to its starting position x=A
so it travels a distance of A+A+A+A=4A
In the macroscopic world, you know that you can hear but cannot see around corners. Under what conditions does light bend around corners (i.e. diffract) ? Explain why sound diffracts easily around a classroom door. 5. Suppose you added to the single slit an identical slit a distance d=0.25mm away from the first. Draw the resulting interference pattern you might expect on the same screen. What happens when we increase the distance between slits ? What happens in the limit that d becomes arbitrarily large?
Answer:
a much larger slit, the phenomenon of Sound diffraction that slits for light.
this is a series of equally spaced lines giving a diffraction envelope
Explanation:
The diffraction phenomenon is described by the expression
d sin θ = m λ
Where d is the distance of the slit, m the order of diffraction that is an integer and λ the wavelength.
For train the diffraction phenomenon, the d / Lam ratio is decisive if this relation of the gap separation in much greater than the wavelength does not reduce the diffraction phenomenon but the phenomena of geometric optics.
The wavelength range for visible light is 4 10⁻⁷ m to 7 10⁻⁷ m. The wavelength range for sound is 17 m to 1.7 10⁻² m. Therefore, with a much larger slit, the phenomenon of Sound diffraction that slits for light.
When we add a second slit we have the diffraction of each one separated by the distance between them, when the integrals are made we arrive at the result of the interference phenomenon, a this is a series of equally spaced lines giving a diffraction envelope
When I separate the distance between the two slits a lot, the time comes when we see two individual diffraction patterns
The center of mass of a pitched baseball or
radius 3.91 cm moves at 33.6 m/s. The ball
spins about an axis through its center of mass
with an angular speed of 52.1 rad/s.
Calculate the ratio of the rotational energy
to the translational kinetic energy. Treat the
ball as a uniform sphere.
The ratio between rotational energy and translational kinetic energy is [tex]1.47\cdot 10^{-3}[/tex]
Explanation:
The translational kinetic energy of the ball is given by:
[tex]KE_t = \frac{1}{2}mv^2[/tex]
where
m is the mass of the ball
v is the speed of the ball
The rotational kinetic energy of the ball is given by
[tex]KE_r = \frac{1}{2}I\omega^2[/tex]
where
I is the moment of inertia
[tex]\omega[/tex] is the angular speed
The moment of inertia of a solid sphere through its axis is given by
[tex]I=\frac{2}{5}mR^2[/tex]
where
m is the mass of the ball
R is its radius
Substituting into the previous equation,
[tex]KE_r = \frac{1}{2}(\frac{2}{5}mR^2)\omega^2 = \frac{1}{5}mR^2 \omega^2[/tex]
The ratio between the two energies is
[tex]\frac{KE_r}{KE_t}=\frac{\frac{1}{5}mR^2 \omega^2}{\frac{1}{2}mv^2}=\frac{2R^2 \omega^2}{5v^2}[/tex]
And substituting:
R = 3.91 cm = 0.0391 m
v = 33.6 m/s
[tex]\omega=52.1 rad/s[/tex]
we find:
[tex]ratio = \frac{2(0.0391)^2(52.1)^2}{5(33.6)^2}=1.47\cdot 10^{-3}[/tex]
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The ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere should be [tex]1.47.10^-3[/tex].
Calculation of the ratio:Since
The translational kinetic energy of the ball should be
[tex]KE_t = 1/2mv^2[/tex]
here
m is the mass of the ball
v is the speed of the ball
Now
The rotational kinetic energy of the ball should be
[tex]KE_r = 1/2Iw^2[/tex]
Here
I is the moment of inertia
w is the angular speed
Now
The moment of inertia of a solid sphere through its axis should be
[tex]I = 2/5mR^2[/tex]
Here
m is the mass of the ball
R is its radius
So,
[tex]KE_r = 1/2(2mR^2)w^2\\\\= 1/5mR62w^2[/tex]
Now
the ratio be
[tex]KE_r/KE_t = (1/2mR^2w^2) / (1/2mv^2)\\\\= (2R^2 w^2) / (5v^2)\\\\= 290.0391)^2 (52.1)^2 / 5(33.6)^2\\\\= 1.47*10^-3[/tex]
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A 477 g portion of soup is heated in a microwave oven from 25°C to 90°C, using radiation with a wavelength of 1.55 × 10⁻² m. Assuming the heat capacity of the soup is the same as water (4.18 J/°C･g), how many photons are absorbed by the soup?
To solve this problem we will use the heat transfer equations, to determine the amount of heat added to the body. Subsequently, through the energy ratio given by Plank, we will calculate the energy of each of the photons. The relationship between total energy and unit energy will allow us to determine the number of photons
The mass of water in the soup is 477g
The change in temperate is
[tex]\Delta T = (90+273K)-(25+273K) = 65K[/tex]
Use the following equation to calculate the heat required to raise the temperature:
[tex]q = mc\Delta T[/tex]
Here,
m = Mass
c = Specific Heat
[tex]q = (477)(4.184)(65)[/tex]
[tex]q = 129724.92J[/tex]
The wavelength of the ration used for heating is [tex]1.55*10^{-2}m[/tex]
The number of photons required is the rate between the total energy and the energy of each proton, then
[tex]\text{Number of photons} = \frac{\text{Total Energy}}{\text{Energy of one Photon}}[/tex]
This energy of the photon is given by the Planck's equation which say:
[tex]E = \frac{hc}{\lambda}[/tex]
Here,
h = Plank's Constant
c = Velocity of light
[tex]\lambda =[/tex] Wavelength
Replacing,
[tex]E = \frac{(6.626*10^{-34})(3*10^8)}{1.55*10^{-2}}[/tex]
[tex]E = 1.28*10^{-23}J[/tex]
Now replacing we have,
[tex]\text{Number of photons} = \frac{82240.7}{1.28*10^{-23}}[/tex]
[tex]\text{Number of photons} = 6.41*10^{27}[/tex]
Therefore the number of photons required for heating is [tex]6.41*10^{27}[/tex]
A 4-A current is maintained in a simple circuit with a total resistance of 2 Ω. How much energy is dissipated in 3 seconds?
A) 3 J
B) 6 J
C) 12 J
D) 24 J
E) 96 J
Answer: Energy dissipated E = 96J
Explanation:
Given:
Current I = 4A
Resistance R = 2 Ohms
Time t = 3 seconds
The energy dissipation in an electric circuit can be derived from the equation below:
E = IVt ....1
Where;
I = current, V = Voltage (potential difference), t= time and E = energy dissipated
But we know that;
V = I×R .....2
Substituting equation 2 to 1, we have
E = IVt = I(I×R)t = I^2(Rt)
Substituting the values of I,R and t
E = 4^2 × 2 ×3 = 96J
Energy dissipated E = 96J
Final answer:
The energy dissipated in the circuit is 96J.
Explanation:
To calculate the energy dissipated in a circuit, you need to use the formula P = IV, where P is power, I is current, and V is voltage. In this case, the current is 4A and the total resistance is 2Ω. Using Ohm's law (V = IR), we can find the voltage as V = I * R = 4A * 2Ω = 8V. Now, we can calculate the power dissipated as P = IV = 4A * 8V = 32W. Lastly, to find the energy dissipated, we multiply the power by the time, so 32W * 3s = 96J.
Make a prediction on how two A-tapes interact. Will they attract, repel or exert no force on each other?
Answer:
The two tapes will either attract or repel each other, depending on the nature of interaction. This is explained below:
Explanation:
When you rip the two pieces of tape off the table, there is a tug-of-war for electric charges between tape and table. The tape either steals negative charges (electrons) from the table or leaves some of its own negative charges behind, depending on what the table is made of (a positive charge doesn’t move in this situation). In any case, both pieces of tape end up with the same kind of charge, either positive or negative. Since like charges repel, the pieces of tape repel each other.
When the tape sandwich is pulled apart, one piece rips negative charges from the other. One piece of tape therefore has extra negative charges. The other piece, which has lost some negative charge, now has an overall positive charge. Because opposite charges attract, the two pieces of tape attract each other.
A 400 kg satellite orbits the moon at a height of 2000 km above the moon at a speed of 946 m/s. The speed of a 800 kg satellite orbiting the moon at 2000 km above the moon is:
To solve this problem we will apply the concepts related to the balance of Forces, in this case the centripetal force of the body must be equal to the gravitational force exerted by the moon on it.
The gravitational force is given by the function
[tex]F_g = \frac{GmM}{r^2}[/tex]
Here
G = Gravitational Universal constant
M = Mass of the planet
m = Mass of the satellite
r = Radius(orbit)
Now the centripetal force is given as
[tex]F_c =\frac{mv^2}{r}[/tex]
Here
m = mass of satellite
v = Velocity of satellite
r = Radius (orbit)
Since there must be balance for the satellite to remain in the orbit
[tex]F_c = F_g[/tex]
[tex]\frac{mv^2}{r} = \frac{GmM}{r^2}[/tex]
[tex]v^2= \frac{GM}{r}[/tex]
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
The velocity depends on the mass of the planet and the orbit, and not on the mass, so if the orbit is maintained, the velocity will be the same: 946m/s
The period
T
of a simple pendulum is the amount of time required for it to undergo one complete oscillation. If the length of the pendulum is
L
and the acceleration of gravity is
g
,
then
T
is given by
T
=
2
π
L
p
g
q
.
Find the powers
p
,
q
required for dimensional consistency.
Answer: p = - {1/2} , q = {1/2}
Explanation: The frequency of oscillation of a pendulum is given as
F = 1/2π *√{l/g}
Where √ is square root
l is lenght
g is acceleration due to gravity
But
F = 1/T
Where T is the period of Oscillation
Making T subject of formula we have
T= 1/F
T = 2π√{g/l}
Here the power on l is -[1/2]= p
Also,
Power on g is 1/2 =q
All because of the square root.
One day when you come into physics lab you find several plastic hemispheres floating like boats in a tank of fresh water. Each lab group is challenged to determine the heaviest rock that can be placed in the bottom of a plastic boat without sinking it. You get one try. Sinking the boat gets you no points, and the maximum number of points goes to the group that can place the heaviest rock without sinking. You begin by measuring one of the hemispheres, finding that it has a mass of 23 g and a diameter of 8.4 cm . What is the mass of the heaviest rock that, in perfectly still water, won't sink the plastic boat?
To solve this problem we will first proceed to find the volume of the hemisphere, from there we will obtain the mass of the density through the relation of density. Finally the mass of the stone will be given between the difference in the mass given in the statement and the one found, that is
The volume of a Sphere is
[tex]V = \frac{4}{3} \pi r^3[/tex]
Then the volume of a hemisphere is
[tex]V =\frac{1}{2} \frac{4}{3} \pi r^3[/tex]
With the values we have that the Volume is
[tex]V =\frac{1}{2} \frac{4}{3} \pi (8.4/2)^3[/tex]
[tex]V = 155.17cm^3[/tex]
Density of water is
[tex]\rho = 1g/cm^3[/tex]
And we know that
[tex]\text{Mass of water displaced} = \text{Density of water}\times \text{Volume of hemisphere}[/tex]
[tex]m = 1g/cm^3 * 155.17cm^3[/tex]
[tex]m = 155.17g[/tex]
So the net mass is
[tex]\Delta m = m_s-m_w[/tex]
[tex]\Delta m = 155.17-23[/tex]
[tex]\Delta m = 132.17g[/tex]
Therefore the mass of heaviest rock is 132.17g or 0.132kg
The mass of the heaviest rock that, in perfectly still water, won't sink the plastic boat is 155 g.
We know that the density of water = 1 g/cm^3
Volume of the hemisphere = 2/3 πr^2
When diameter = 8.4 cm, radius = 4.2 cm
So, V = 2/3 × 3.14 × (4.2)^3
V = 155 cm^3
Volume of hemisphere = volume of water displaced = 155 cm^3
Mass of water displaced = 155 g
Since the solid displaces its own mass of water, the mass of the heaviest rock that, in perfectly still water, won't sink the plastic boat is 155 g.
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A student who grew up in a tropical country and is studying in the United States may have no experience with static electricity sparks and shocks until their first American winter. Explain.
Explanation:
Water in the air (humidity) helps to dissipate static charge that builds up. If the air is very dry, the charge can't dissipate, so it builds up until there is enough to spark.
Tropical countries are typically more humid than the United States, but I guess that depends on where you are in the US.
A student's first experience with static electricity in an American winter is due to the dry air, which allows for greater accumulation and discharge of electrical charges, unlike in humid tropical climates.
This is because static electricity is more prevalent in cold, dry environments.
In tropical countries, the air tends to have higher humidity, which allows electrical charges to dissipate more easily. Moist air is a better conductor of electricity, thereby reducing the likelihood of a significant buildup of static charge.In contrast, during an American winter, the air is typically cold and dry. Dry air is a poor conductor, meaning that electrical charges are more likely to accumulate on surfaces and objects, leading to more frequent static electricity shocks. When you walk on a rug or take off a woollen sweater in such conditions, electrons can be transferred between your body and the surfaces, creating an imbalance.As a result, when you touch a metal object like a doorknob, the built-up charge is suddenly released, creating a spark and a shock. Even though the spark may carry a few hundred watts of power, it happens so quickly and involves such a small amount of current that it doesn't cause injury. The brief duration and low current ensure that the shock is felt but not harmful.In summary, the primary reason a student may experience static electricity for the first time in an American winter is due to the dry air conditions that favour the accumulation and sudden discharge of electric charges.
During an ice show a 60 kg skater leaps into the air and is caught by an initially stationary 75.0 kg skater.a. What is their final velocity assuming negligible friction and that the 60.0-kg skater's original horizontal velocity is 4.00 m/s?b. How much kinetic energy is lost?
Answer:
(a). The final velocity is 1.78 m/s.
(b). The lost kinetic energy is 266.13 J.
Explanation:
Given that,
Mass of first skater = 60 kg
Mass of second skater = 75.0 kg
Initial velocity = 4.00 m/s
(a). We need to calculate the final velocity
Using conservation of momentum
[tex]m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v[/tex]
Put the value into the formula
[tex]60\times4.00+75.0\times0=(60+75.0)v[/tex]
[tex]v=\dfrac{60\times4.00}{60+75}[/tex]
[tex]v=1.78\ m/s[/tex]
The final velocity is 1.78 m/s.
(b). We need to calculate the lost kinetic energy
Using formula of kinetic energy
[tex]\Delta E=E_{2}-E_{1}[/tex]
[tex]\Delta E=\dfrac{1}{2}(m_{1}u_{1}^2+m_{2}u_{2}^2)-\dfrac{1}{2}(m_{1}+m_{2})v^{2}[/tex]
[tex]\Delta E=\dfrac{1}{2}(60\times4^2)-\dfrac{1}{2}\times(60+75)\times1.78^2[/tex]
[tex]\Delta E=266.13\ J[/tex]
Hence, (a). The final velocity is 1.78 m/s.
(b). The lost kinetic energy is 266.13 J.
a. The final velocity should be considered as the 1.78 m/s.
b. The lost kinetic energy should be considered as the 266.13 J.
Conservation of momentum:Since
Mass of first skater = 60 kg
Mass of second skater = 75.0 kg
Initial velocity = 4.00 m/s
a.
We know that
[tex]mm1v1 + m2v2 = (m1 + v1)v\\\\60*4.00 + 75*0 = (60 + 75)v[/tex]
v = 60 + 4/60 + 75
= 1.78 m/s
b.
[tex]E = 1/2(m1v1^2 + m2v2^2) - 1/2(m1 + vm2)v^2\\\\= 1/2(60*4^2) - 1/2 * (60 + 75)*1.78^2[/tex]
= 266.13 J
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In an automobile collision, how does an airbag lessen the blow to the passenger? Assume as a result of the collision, the passenger stops.a.The air bag decreases the momentum change of the passenger in the collision.b.
During the collision, the force from the air bag isgreater than would be the force from the wind-shield or dashboard so the passenger cannot hit the hard objects.c.The stopping impulse is the same for either the hard objects or the airbag. Unlike the windshieldor dashboard, the air bag gives some increasing the time for the slowing process and thus de-creasing the average force on the passenger.d.The airbag is there to insure the seatbelt holds
Answer:
c.The stopping impulse is the same for either the hard objects or the airbag. Unlike the windshield or dashboard, the air bag gives some increasing the time for the slowing process and thus decreasing the average force on the passenger
Explanation:
As soon as the vehicle collides the air bags open up almost instantly within a fraction of seconds.The air bag reduces the force of impact by providing the longer time for the change in momentum in accordance with the Newton's second law and distributes the force over a larger area of impact therefore reducing the pressure.The impulse in each case remains the same because when the air bags are engaged in the process the force is lesser but the time involved is more and we know that impulse is the product of force and time.The stopping impulse is the same for either the hard objects or the airbag.
Unlike the windshield or dashboard, the airbag gives some increasing the
time for the slowing process and thus decreasing the average force on the
passenger.
Air bags are put in automobiles ion order to prevent life threatening injuries
such as brain damage to individuals. The air bag contains air helps
to increase the time for the slowing process.
This thereby prevents the average force on the passenger and prevents
injuries which should have occurred.
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Knowing that α= 60° and that boom AC exerts on pin C a force directed along line AC, determine (a)the magnitude of t hat force, (b) the tension in cable BC.Figure:Cable BC is making an angle alpha with fixed support.
Answer:
Tbc = 230.69 N ; Fac = 172.31 N
Explanation:
Sum of forces in y direction:
[tex]T_{BC} * sin (35) = 300*sin (70) + F_{AC}*sin (60) .... Eq 1\\[/tex]
Sum of forces in x direction:
[tex]T_{BC} * cos (35) + F_{AC}*cos (60)= 300*cos (70) .... Eq 1\\[/tex]
Solving Eq 1 and Eq 2 simultaneously:
[tex]T_{BC} = 281.9077862 + \sqrt{3} / 2 * F_{AC}\\\\F_{AC} (1.736868124) = 300*cos (70) - 491.4912266*cos (35)\\\\F_{AC} = - \frac{300}{1.736868124}\\\\F_{AC} = - 172.73 N\\\\T_{BC} = 230.69 N[/tex]
Answer: Tbc = 230.69 N ; Fac = 172.31 N
(a) The tension in the cable at AC is -200.67 N.
(b) The tension in the cable at BC is 328.99 N.
Forces resolved along y-axisThe sum of the forces in y-direction is calculated as follows;
T(BC)sin(35) = 400 x sin(65) + F(AC) sin(60) --- (1)
Forces resolved along x-axisT(BC)cos(35) + F(AC) cos(60) = 400 x cos(65) ---(2)
From equation(1);
[tex]T_{BC} = \frac{400 \times sin(65) \ + \ F_{AC} sin(60)}{sin(35)} \\\\T_{BC} = 632 + 1.51F_{AC}[/tex]
From equation (2);
0.82T(BC) + 0.5F(AC) = 169.1
[tex]0.82(632 + 1.51F_A_C) + 0.5F_A_C= 169.1\\\\518.24 + 1.24F_A_C + 0.5F_A_C = 169.1\\\\F_A_C = \frac{-349.14}{1.74} \\\\F_A_C = -200.67 \ N[/tex]
T(BC) = 632 + 1.51(-200.67)
T(BC) = 328.99 N
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Mass–spring systems are used as tuned mass dampers to diminish the vibrations of the balconies of a performing arts center. The oscillation frequency of the TMD is 6.85 Hz, the oscillating mass is 142 kg, and the oscillation amplitude is 4.86 cm.
A) What is the spring constant?
B) What is the maximum speed of the mass?
C) What is the maximum accelerations of the mass?
Answer:
A) [tex]k=2.63*10^{5} N/m[/tex].
B)[tex]v=2.10m/s[/tex]
C)[tex]a=90.0m/s^{2}[/tex]
Explanation:
This problem is a simple harmonic motion problem. The equation of motion for the SHM is:
[tex]\frac{d^{2}x}{dt^{2}}+\omega^{2}x=0[/tex],
where x is the displacement of the mass about its point of equilibrium, t is time, and [tex]\omega[/tex] is the angular frequency.
A)
First, we need to remember that
[tex]\omega^{2}=\frac{k}{m}[/tex],
where k is the spring constant, and m is the mass.
From here we can simply solve for k, so
[tex]k=\omega^{2}m[/tex].
Now, we need to make use of an equation that relates the frequency and angular frequency. The equetion is
[tex]\omega=2\pi \nu[/tex],
where [tex]\nu[/tex] is the frequency. This leads us to
[tex]k=(2\pi \nu)^{2}m[/tex],
[tex]k=142(2*6.85*\pi)^{2}[/tex],
[tex]k=2.63*10^{5} N/m[/tex],
B) In simple harmonic motion, the velocity behaves as follow:
[tex]v=\omega Acos(\omega t)[/tex] (this is obtained by solving the equation of motion of the mass for the displacement x and take the derivative),
where A is the amplitude of the motion. Since we want the maximum value for the speed, we make [tex]cos(\omega t)=1[/tex] (this because cosine function goes from -1 to 1). With this, the maximum speed is simply
[tex]v = \omega A\\v=(2\pi \nu)A\\v=(2*6.85*\pi)*0.0486\\v=2.10m/s[/tex]
C) Here we are going to use the equation of motion of SHM
[tex]\frac{d^{2}x}{dt^{2}}+\omega^{2}x=0[/tex],
we know that
[tex]a=\frac{d^{2}x}{dt^{2}}[/tex] , where a is the acceleration,
[tex]a+\omega^{2}x=0\\a=-\omega^{2}x[/tex]
in this case, x goes from -A to A, so for a to be maximum we need that [tex]x=-A[/tex] ,and we get
[tex]a=-\omega^{2}(-A)\\a=\omega^{2}A\\a=(2\pi \nu)^{2}A\\a=(2*6.85*\pi)^{2}(0.0486)\\a=90.0m/s^{2}[/tex]
One end of rod A is placed in a cold reservoir with a temperature of 5.00°C. The other end is held in a hot reservoir at 85.0°C. Rod A has a length L and a radius r. Rod B is made of the same material as rod A and the ends of rod B are placed in the same reservoirs as rod A. Rod B has a length 2L and a radius 2r. What is the ratio of heat flow through rod A to that through rod B?
Answer:
1 / 2
Explanation:
This problem is a 1 - D steady state heat conduction with only one independent variable (x).
1 - D steady state:
Q = dT / Rc
Q = heat flow
dT = change in temperature between a pair of node
Rc = thermal resistance
Rc = L / k*A
Since in both cases Rod A and Rod B have identical boundary conditions:
dT_a = dT_b
So,
R_a = L / k*(pi*r^2)
R_b = 2L / k*(pi*(2r)^2) = L / k*(2*pi*r^2)
Compute Q_a and Q_b:
Q_a = k * dT *(pi * r^2 * / L)
Q_a = k * dT*(2*pi * r^2 * / L)
Ratio of Q_a to Q_b
Q_a / Q_b = [k * dT *(pi * r^2 * / L)] / [k * dT*(2*pi * r^2 * / L)] = 1 / 2
Final answer:
The ratio of heat flow through rod A to rod B is 1:2 when rod B has double the length and radius of rod A, both rods being made of the same material and subjected to the same temperature difference. This conclusion is derived from Fourier's law of thermal conduction.
Explanation:
The question asks about the comparison of heat flow through two rods of different dimensions but made from the same material and exposed to the same temperature difference. To find the ratio of heat flow through rod A to that through rod B, we use Fourier's law of thermal conduction, which states that the rate of heat transfer through a material is proportional to the negative gradient of temperature and the cross-sectional area of the material, and inversely proportional to the length of the material's path. Mathematically, we write this as Q = (kAΔT) / L, where Q is the heat transfer per unit time, k is the thermal conductivity of the material, A is the cross-sectional area, ΔT is the temperature difference, and L is the length of the material's path.
For rod A, assuming a unit thermal conductivity for simplicity, the rate of heat transfer QA = (kπr2(85.00 - 5.00)) / L. For rod B, with double the radius and length, QB = (kπ(2r)2(85.00 - 5.00)) / 2L. Simplifying these expressions, we find that the ratio of heat flow through rod A to that through rod B is QA/QB = 1/2. Thus, rod A transfers heat at half the rate of rod B under the given conditions.
Knowing the P is 1110 N, determine the tension in the rope if the frame is in equilibrium. Express your answer in kN to the nearest 10 N.
Explanation:
It is given that the value of P is 1110 N. And, for pin connection we have only two connections which are [tex]A_{x}[/tex] and [tex]A_{y}[/tex]. Let T be the tension is rope.
So, [tex]\sum F_{y} = 0[/tex] and [tex]A_{y}[/tex] - 1110 = 0
[tex]A_{y} = 1110 N[/tex]
[tex]\sum F_{x}[/tex] = 0
And, [tex]T - A_{x}[/tex] = 0
T = [tex]A_{x}[/tex]
Also, [tex]\sum M_{A}[/tex] = 0
1110(0.75 + 0.75 + 0.75) - T(0.5 + 0.1) = 0
2497.5 - 0.6T = 0
T = 4162.5 N
= 4.16 kN
Therefore, we can conclude that the tension in the rope if the frame is in equilibrium is 4.16 kN.
In a 5.00 km race, one runner runs at a steady 11.4 km/h and another runs at 14.7 km/h . How long does the faster runner have to wait at the finish line to see the slower runner cross?
Answer:
0.0986 h or 5 minutes 55 seconds.
Explanation:
Speed: This can be defined as the rate of change of distance of a body. The S.I unit of speed is m/s. Speed is a scalar quantity, because it can only be represented by magnitude alone.
Mathematically,
Speed = distance/time.
S = d/t ........................... Equation 1
making t the subject of the equation
t = d/S ......................... Equation 2
Form the question,
Time taken for the faster runner to reach the finish line
t₁ = d/S₁................... Equation 3
Where t₁ = time taken for the faster runner to reach the finish line, d = distance, S₁ = speed of the faster runner.
Given: d = 5.0 km, S₁ = 14.7 km/h.
Substituting into equation 3
t₁ = 5/14.7
t₁ = 0.340 h
Also,
t₂ = d/S₂................... Equation 4
Where t₂ = time taken for the slower runner to reached the finished line, d = distance, S₂ = speed of the slower runner.
Given: d = 5 km, S₂ = 11.4 km/h.
Substitute into equation 4,
t₂ = 5/11.4
t₂ = 0.4386 h.
The time the faster runner have to wait at the finish line to see the slower runner cross = t₂ - t₁ = 0.4386-0.340
The time the faster runner have to wait at the finish line to see the slower runner cross = 0.0986 h = 5 mins 55 s.
A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.42 . The angle between the horizontal and the ladder is θ . The person wants to climb up the ladder a distance of 0.9 m along the ladder from the ladder’s foot. 16.2 kg 0.9 m 2 m θ b µ = 0.42 µ = 0 What is the minimum angle θmin (between the horizontal and the ladder) so that the person can reach a distance of 0.9 m without having the ladder slip? The acceleration of gravity is 9.8 m/s 2
To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.
The forces in the vertical direction would be,
[tex]\sum F_x = 0[/tex]
[tex]f-N_w = 0[/tex]
[tex]N_w = f[/tex]
The forces in the horizontal direction would be,
[tex]\sum F_y = 0[/tex]
[tex]N_f -W =0[/tex]
[tex]N_f = W[/tex]
The sum of Torques at equilibrium,
[tex]\sum \tau = 0[/tex]
[tex]Wdcos\theta - N_wLsin\theta = 0[/tex]
[tex]WdCos\theta = fLSin\theta[/tex]
[tex]f = \frac{Wd}{Ltan\theta}[/tex]
The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore
[tex]f_{max} = \mu W=\frac{Wd}{Ltan\theta}[/tex]
[tex]\theta = tan^{-1} (\frac{d}{\mu L})[/tex]
Replacing,
[tex]\theta = tan^{-1} (\frac{0.9}{0.42*2})[/tex]
[tex]\theta = 46.975\°[/tex]
Therefore the minimum angle that the person can reach is 46.9°