Answer:
Part a)
Part b)
Part c)
Part d)
Explanation:
Part a)
As we know that speed of package is same as that of helicopter in horizontal direction
So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall
So we have
Part b)
Distance from helicopter is same as the distance of free fall
so we will have
Part c)
If helicopter is rising upwards with uniform speed
then final speed of the package after time t is given as
Part d)
distance from helicopter
Answer:
- 3 meter
Explanation:
A dog has started motion from +3 meter. ...(Given)
∴ maximum positive distance = + 3 meter
Magnitude of distance = 3
Maximum negative distance = (-) (magnitude of distance)
Maximum negative distance = (-) (3)
Maximum negative distance = -3 meter
Hence, maximum negative distance is -3 meter.
Kinetic energy is calculated through the equation,
KE = 0.5mv²
At initial conditions,
m₁: KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J
m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J
Due to the momentum balance,
m₁v₁ + m₂v₂ = (m₁ + m₂)(V)
Substituting the known values,
(0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)
V = 0.2977 m/s
The kinetic energy is,
KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
KE = 0.03146 J
The difference between the kinetic energies is 0.0473 J.
Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration, 
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,
Frictional force is given by :
F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.
