The initial kinetic energy of the boat and its rider is
After Sam stops it, the final kinetic energy of the boat+rider is
because its final velocity is zero.
For the law of conservation of energy, the work done by Sam is the variation of kinetic energy of the system:
where the negative sign is due to the fact that the force Sam is applying goes against the direction of motion of the boat.
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Answer:
The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
Explanation:
From the question given above, the following data were obtained:
Height to which the target is located = 50 m
Initial velocity (u) = 20 m/s
To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = 10 m/s²
Maximum height (h) =?
v² = u² – 2gh (since the ball is going against gravity)
0² = 20² – (2 × 10 × h)
0 = 400 – 20h
Collect like terms
0 – 400 = – 20h
– 400 = – 20h
Divide both side by – 20
h = – 400 / – 20
h = 20 m
Thus, the the maximum height to which the cannon ball attained is 20 m.
From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
Answer:
planet that is farthest away is planet X
kepler's third law
Explanation:
For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets
T² = (
a³ = K_s a³
Let's apply this equation to our case
a = ![\sqrt[3]{ \frac{T^2}{K_s} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%20%5Cfrac%7BT%5E2%7D%7BK_s%7D%20%7D)
for this particular exercise it is not necessary to reduce the period to seconds
Plant W
10² = K_s 
a_w = ![\sqrt[3]{ \frac{100}{ K_s} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%20%5Cfrac%7B100%7D%7B%20K_s%7D%20%7D)
a_w = ![\frac{1}{ \sqrt[3]{K_s} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%20%5Csqrt%5B3%5D%7BK_s%7D%20%7D)
4.64
Planet X
a_x = ![\sqrt[3]{ \frac{640^3}{K_s} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%20%5Cfrac%7B640%5E3%7D%7BK_s%7D%20%7D)
a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3
Planet Y
a_y = ![\sqrt[3]{ \frac{80^2}{K_s} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%20%5Cfrac%7B80%5E2%7D%7BK_s%7D%20%20%7D)
a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6
Planet z
a_z = ![\sqrt[3]{ \frac{270^2}{K_s} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%20%5Cfrac%7B270%5E2%7D%7BK_s%7D%20%7D)
a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8
From the previous results we see that planet that is farthest away is planet X
where we have used kepler's third law
