The relevant equation we can use in this problem is:
h = v0 t + 0.5 g t^2
where h is height, v0 is initial velocity, t is time, g is
gravity
Since it was stated that the rock was drop, so it was free
fall and v0 = 0, therefore:
h = 0 + 0.5 * 9.81 m/s^2 * (4.9 s)^2
<span>h = 117.77 m</span>
The motion of the racers might change from the start because the pressure goes up so all the racer wants is to speed up and win, so when the racer first starts he or she is calm because he's not driving yet and when he or she is on his/hers way to he finish line he/she just wants to win and gets under pressure so he speeds up even more and drifts. Your welcome
Answer:

Explanation:
from the ideal gas law we have
PV = mRT


HERE R is gas constant for dry air = 287 J K^{-1} kg^{-1}


We know by ideal gas law



for m_2



WE KNOW
PV = mRT
V, R and T are constant therefore we have
P is directly proportional to mass




Answer:
E. Student 1 is correct, because as θ is increased, h is the same.
Explanation:
Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.
Mathematically:

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.
And for work we have:

where:
displacement of the object
angle between the force and displacement vectors
Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.
- In case of free-fall the angle between the force is and the displacement is zero.
- In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.
So, the work done by the gravitational force is same in the two cases.