Explanation:
Uniform velocity is when an object goes an equal amount of space in an equal amount of time whereas non uniform velocity is when the object covers an unequal amount of distance in an equal amount of time.
Answer:
56.86153 N
Explanation:
t =Time taken
F = Force
Power

Work done

The magnitude of the force that is exerted on the handle is 56.86153 N
Answer: C.
Explanation:
For a parallel-plate capacitor where the distance between the plates is d.
The capacitance is:
C = e*A/d
You can see that the distance is in the denominator, then if we double the distance, the capacitance halves.
Now, the stored energy can be written as:
E = (1/2)*Q^2/C
Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:
E = (1/2)*Q^2*d/(e*A)
e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.
Then we can define:
K = (1/2)*Q^2/(e*A)
And now we can write the energy as:
E = K*d
Then the energy is proportional to the distance between the plates, this means that if we double the distance, we also double the energy.
Explanation:
Initial energy = final energy + work done by friction
PE = PE + KE + W
mgH = mgh + 1/2 mv² + W
(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000
v = 22.1 m/s
Without friction:
PE = PE + KE
mgH = mgh + 1/2 mv²
(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²
v = 23.4 m/s
a. 0.5 T
- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position
- The period T is the time the system takes to complete one oscillation
During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.
So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

and solving for t we find

b. 1.25T
Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that
- the mass takes a time of 1 T to cover a distance of 4A
we can set the following proportion:

And by solving for t, we find
