B. The apple from the bottom will hit the ground earlier. This is because an increase in height causes an increase in the time that the object will fall, and therefore will affect the final velocity of the falling object. Moreover, the reduction in velocity due to friction from the air should also be considered.
Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant
The work done in stretching the rubber band from x = 0 to x = L is
![W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7BL%7D%20Fdx%20%3D%20%5Cint_%7B0%7D%5E%7BL%7Dax%20%5C%2C%20dx%20%3D%20%5Cfrac%7Ba%7D%7B2%7D%20%20%5Bx%5E%7B2%7D%20%5D_%7B0%7D%5E%7BL%7D%20%3D%20%20%5Cfrac%7BaL%5E%7B2%7D%7D%7B2%7D%20)
Answer:
Answer:
1470 W
Explanation:
Power: This can be defined as the rate at which work is done or energy is used up. The S.I unit of power is Watt (W).
The expression for power is given as,
P = Energy/time
P = mgh/t ...................... Equation 1
Where P = power, m = mass, h = height, t = time, g = acceleration due to gravity.
Given: m = 75 kg, g =9.8 m/s², h = 1 m, t = 1 s.
Substitute into equation 1
P = (75×1×9.8)/1
P = 735 W.
From the above,
1 hp = 735 W
2 hp = (2×735) W
2 hp = 1470 W.
Hence 2 hp = 1470 W
Here is the rule for see-saws here on Earth, and there is no reason
to expect that it doesn't work exactly the same anywhere else:
(weight) x (distance from the pivot) <u>on one side</u>
is equal to
(weight) x (distance from the pivot) <u>on the other side</u>.
That's why, when Dad and Tiny Tommy get on the see-saw, Dad sits
closer to the pivot and Tiny Tommy sits farther away from it.
(Dad's weight) x (short length) = (Tiny Tommy's weight) x (longer length).
So now we come to the strange beings on the alien planet.
There are three choices right away that both work:
<u>#1).</u>
(400 N) in the middle-seat, facing (200 N) in the end-seat.
(400) x (1) = (200) x (2)
<u>#2).</u>
(200 N) in the middle-seat, facing (100 N) in the end-seat.
(200) x (1) = (100) x (2)
<u>#3).</u>
On one side: (300 N) in the end-seat (300) x (2) = <u>600</u>
On the other side:
(400 N) in the middle-seat (400) x (1) = 400
and (100 N) in the end-seat (100) x (2) = 200
Total . . . . . . . . . . . . <u>600</u>
These are the only ones to be identified at Harvard . . . . . . .
There may be many others but they haven't been discarvard.
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
