Answer:
If the force remains the same, the acceleration would decrease
Explanation:
According to Newton's second law, the acceleration of an object is given by
![a=\frac{F}{m}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF%7D%7Bm%7D)
where
F is the force applied to the object
m is the mass of the object
As we see from the formula, the acceleration a is inversely proportional to the mass, m. Therefore, if the force F remains constant, this means that if the mass of the skateboarder increases, then the acceleration will decrease.
Can I still get 5 points bc u already figured it out
Answer:
Power=?,time=15,mass=3,height=50,g=10.
Power=mgh/t......3×10×50=1500/15=100Watts
Answer:
a). Determine the magnitude of the gravitational force exerted on each by the earth.
Rock: ![F = 49.06N](https://tex.z-dn.net/?f=F%20%3D%2049.06N)
Pebble: ![F = 29.44N](https://tex.z-dn.net/?f=F%20%3D%2029.44N)
(b)Calculate the magnitude of the acceleration of each object when released.
Rock: ![a =9.8m/s^{2}](https://tex.z-dn.net/?f=a%20%3D9.8m%2Fs%5E%7B2%7D)
Pebble: ![a =9.8m/s^{2}](https://tex.z-dn.net/?f=a%20%3D9.8m%2Fs%5E%7B2%7D)
Explanation:
The universal law of gravitation is defined as:
(1)
Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.
<em>Case for the rock </em>
<em>:</em>
m1 will be equal to the mass of the Earth
and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth
.
![F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}](https://tex.z-dn.net/?f=F%20%3D%20%286.67x10%5E%7B-11%7Dkg.m%2Fs%5E%7B2%7D.m%5E%7B2%7D%2Fkg%5E%7B2%7D%29%5Cfrac%7B%285.972x10%5E%7B24%7D%20Kg%29%285.0%20Kg%29%7D%7B%286371000%20m%29%5E%7B2%7D%7D)
Newton's second law can be used to know the acceleration.
![F = ma](https://tex.z-dn.net/?f=F%20%3D%20ma)
(2)
![a =9.8m/s^{2}](https://tex.z-dn.net/?f=a%20%3D9.8m%2Fs%5E%7B2%7D)
<em>Case for the pebble </em>
<em>:</em>
![F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}](https://tex.z-dn.net/?f=F%20%3D%20%286.67x10%5E%7B-11%7Dkg.m%2Fs%5E%7B2%7D.m%5E%7B2%7D%2Fkg%5E%7B2%7D%29%5Cfrac%7B%285.972x10%5E%7B24%7D%20Kg%29%283.0%20Kg%29%7D%7B%286371000%20m%29%5E%7B2%7D%7D)
![a =9.8m/s^{2}](https://tex.z-dn.net/?f=a%20%3D9.8m%2Fs%5E%7B2%7D)