Answer:
b. The clock in orbit runs slow. The clock in Colorado runs slow.
Explanation:
b. The clock in orbit runs slow. The clock in Colorado runs slow. This is because of the fact that as the gravity weakens, time moves faster, hence, the answer is b.
Answer: The clock in Colorado runs slow
Explanation: I got it correct on the HW. It should be a separate answer from B though.
A teacher does a demonstration in front of a class. For the demonstration, a motion sensor is placed at the end of a long, horizontal track of negligible friction. A force sensor is used to pull a cart of known mass along the track, and the motion sensor tracks the motion of the cart. For several positions of the cart, a computer records the reading on the force sensor and the reading on the motion sensor for both the speed of the cart and the distance the cart moves away from the motion sensor. Based on only these data, which of the following scientific hypotheses could a student test?
a. The force exerted on the cart is directly proportional to the time the cart has moved.
b. The gravitational acceleration for an object on Earth is constant near Earth's surface.
с. The net work done on the cart is equal to the change in kinetic energy of the cart.
d. As the force the sensor exerts on the cart increases, so does the acceleration of the cart.
e. Earth's gravitational force is a conservative force.
Answer:
с. The net work done on the cart is equal to the change in kinetic energy of the cart.
Explanation:
In the context, a motion sensor detects the motion of the cart where a cart is used to move using a force sensor that is attached to the cart of certain mass. As the cart moves with a speed at a particular time, the computer records both the readings on the force sensor as well as the motion sensor. From these readings the is can be concluded that the work done on the cart is same as the change in the kinetic energy in the cart.
One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is at rest as it hangs from the string. When the object is raised to a position X that is a height H above its lowest point and released from rest, the object undergoes simple harmonic motion. When the object passes through the equilibrium position Y, it has a speed v0.
What methods could a student use to determine the total mechanical energy E at position Y, and why?
Answer:
v₀ = [tex]\sqrt{2gH}[/tex]
to determine this speed experimentally, the student must measure the height of the body as a function of time and with equation (1) he can find the speed for each point of interest
Explanation:
In this internal exercise the student must use the conservation of mechanical energy,
Starting point. Highest point of the trajectory
Em₀ = U = m g H
Point of interest. Point at height Y
[tex]Em_{f}[/tex] = K + U = ½ m v² + m g Y
energy is conserved
Em₀ = Em_{f}
m g H = ½ m v² + m g Y
v² = 2 g (H -Y) (1)
in this case they indicate that Y is the equilibrium position whereby Y = 0 and the velocity is v = v₀
v₀ = [tex]\sqrt{2gH}[/tex]
Therefore, to determine this speed experimentally, the student must measure the height of the body as a function of time and with equation (1) he can find the speed for each point of interest
g A 240-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s
Answer:
339.3 N
Explanation:
First, we start by converting the units.
1 rev/s = 2π rad/s, so
0.6 rev/s = 2π * 0.6 rad/s
0.6 rev/s = 1.2π rad/s
0.6 rev/s = 3.77 rad/s
Now we apply the equation of motion,
W(f) = w(o) + αt
3.77 = 0 + α * 2
3.77 = 2α
α = 3.77/2
α = 1.885 rad/s²
Torque = I * α
Torque = F * r
This means that
I * α = F * r, where I = 1/2mr²
Substituting for I, we have
1/2mr²α = F * r, making F the subject of formula, we have
F = 1/2mrα, then we substitute for the values
F = 1/2 * 240 * 1.5 * 1.885
F = 678.6 / 2
F = 339.3 N
You push a manual lawn mower across the lawn at constant speed. What is the value of the coefficient of friction between the mower and the grass
Answer:
0.27Explanation:
The question is incomplete. Here is the complete question:
You are pushing a 13.3 kg lawn mower across the lawn with a force of 200 N. What is the value of the coefficient of friction between the mower and the grass if the mower moves with a constant velocity? The force is applied downward at an angle of 65° with the horizontal.
According to Newton's second law of motion:
[tex]\sum F_x= ma_x\\F_{app} - F_f = ma_x\\[/tex]
[tex]F_f = \mu R\\[/tex]
[tex]F_{app} - \mu Rcos \theta = ma_x[/tex]
Fapp is the applied force = 200N
Ff is the frictional force
[tex]\mu[/tex] is the coefficient of friction between the mower and the grass
R is the reaction
m is the mass of the object
ax is the acceleration
Given
R = mg = 13.3*9.8
R = 130.34N
m = 13.3kg
ax = 0m/s² (constant velocity)
Fapp = 200N
[tex]\theta = 65^0[/tex]
Substitute the given parameters into the formula and get the coefficient of friction as shown;
Recall that: [tex]F_f = \mu R\\[/tex]
[tex]\mu = \frac{F_f}{R}\\\mu = \frac{F_{x}cos65}{F_y+W} \\\mu =\frac{ 200cos65}{200sin65+13.3(9.8)}\\\mu = \frac{84.52}{181.26+130.34}\\\mu = \frac{84.52}{311.6}\\\mu = 0.27[/tex]
Hence the coefficient of friction between the mower and the grass is 0.27
Velocity ratio of single movable pulley is 2 when Velocity ratio of single fixed pulley is 1. why
Answer:
Velocity ratio of single pulley = 2
Explanation:
Given:
Tension T
Distance = T + T = 2T
Find:
Velocity ratio of single pulley
Computation:
Velocity ratio of single pulley = Total distance / Distance effort
Velocity ratio of single pulley = 2T / T
Velocity ratio of single pulley = 2