Answer: 0m/s²
Explanation:
Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane
According newton's law of motion
Summation of forces along the plane = mass × acceleration
Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane
Ff = nR where
n is coefficient of friction = tan(theta)
R is normal reaction = Wcos(theta)
Fm = Wsin(theta)
Substituting in the formula of newton's first law we have;
Fm-Ff = ma
Wsin(theta) - nR = ma
Wsin(theta) - n(Wcos(theta)) = ma... 1
Given
W = 562N, theta = 30°, n = tan30°, m = 56.2kg
Substituting in eqn 1,
562sin30° - tan30°(562cos30°) = 56.2a
281 - 281 = 56.2a
0 = 56.2a
a = 0m/s²
This shows that the trunk is not accelerating
Because the molecules that move freely begin to compact closer together, with less heat, means less molecular activity.
Hello!

Use the equation:

Where:
m = mass of the object (kg)
g = acceleration due to gravity (≈9.8 m/s)
h = height above ground (m)
Plug the given values into the equation:
PE = 7500 · 9.8 · 100
PE = 7,350,000 Joules.
Answer:
just before landing the ground
Explanation:
Let the velocity of projection is u and the angle of projection is 30°.
Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.
initial horizontal component of velocity, ux = u Cos 30
initial vertical component of velocity, uy = u Sin 30
Time of flight is given by

Final horizontal component of velocity, vx = ux = u Cos 30
Let vy is teh final vertical component of velocity.
Use first equation of motion
vy = uy - gT


vy = - u Sin 30
The magnitude of final velocity is given by


v = u
Thus, the velocity is same as it just reaches the ground.