Answer:
The options are not shown, so let's derive the relationship.
For an object that is at a height H above the ground, and is not moving, the potential energy will be:
U = m*g*H
where m is the mass of the object, and g is the gravitational acceleration.
Now, the kinetic energy of an object can be written as:
K = (1/2)*m*v^2
where v is the velocity.
Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.
Uinitial = Kfinal.
m*g*H = (1/2)*m*v^2
v^2 = 2*g*H
v = √(2*g*H)
So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.
Answer:
3.06m/s² to the east
Explanation:
Given parameters:
Mass of car = 2.5 x 10³kg
Force acting on the car = 7.65 x 10³N
Unknown:
Acceleration of the car = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
Acceleration = 
= 
= 3.06m/s² to the east
The angle between 2 o'clock and 12 o'clock is referred to as the angle of twist. The angle between the planes of maximum shear which is bisected by the axis of greatest compression is angle of shear.
300 miles / 6 hours = 50 miles per hour
