Answer:
g' = g/9 = 1.09 m/s²
Explanation:
The magnitude of free fall acceleration at the surface of earth is given by the following formula:
g = GM/R² ----- equation 1
where,
g = free fall acceleration
G = Universal Gravitational Constant
M = Mass of Earth
R = Distance between the center of earth and the object
So, in our case,
R = R + 2 R = 3 R
Therefore,
g' = GM/(3R)²
g' = (1/9) GM/R²
using equation 1:
g' = g/9
g' = (9.8 m/s)/9
<u>g' = 1.09 m/s²</u>
At the highest point of the trajectory the vertical component will have its zero velocity, and the descent caused by the force of gravity will begin.
Since the ball is thrown with a certain speed, the vertical component reaches its highest point (upwards), until returning to the receiver who will receive the ball with the same vertical component but in the opposite direction (downwards).
Therefore the vertical component will have its highest value at launch.
The kinetic energy of an object is given by
KE = 0.5mv²
where m is the mass and v is the velocity.
To calculate the change in kinetic energy...
Initial KE:
KEi = 0.5mVi²
where Vi is the initial velocity.
Final KE:
KEf = 0.5mVf²
where Vf is the final velocity.
ΔKE = KEf - KEi
ΔKE = 0.5mVi² - 0.5mVf²
ΔKE = 0.5m(Vf²-Vi²)
Given values:
m = 16kg
Vi = 25m/s
Vf = 20m/s
Plug in the given values and solve for ΔKE:
ΔKE = 0.5×16×(20²-25²)
ΔKE = -1800J
Answer: (c) 2000 N
Explanation:
Given Data :
▪ Initial velocity = zero ( body is at rest)
▪ Distance travelled = 100m
▪ Final kinetic energy = 200000J
To Find :
▪ Resultant force acting on the car.
Therefore:
W = F × d = ΔK ----------------- eq 1.
where,
W = work done
F = applied force
d = distance
ΔK = change in kinetic energy
Calculation :
→ F × d = Kf - Ki ----------------- eq 2.
Where:
Kf = Final kinetic energy = 200000
Ki = initial kinetic energy = 0
Substituting our values into the formula from equation (2)
→ F × 100 = 200000 - 0
→ F = 200000/100
→ F = 2000N
F = m * a = 60kg * 2 m/s^2 = 120 N
