Answer:
v = 1.28 m/s
Explanation:
Given that,
Maximum compression of the spring, 
Spring constant, k = 800 N/m
Mass of the block, m = 0.2 kg
To find,
The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.
Solution,
When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,
Initial energy = Final energy
Substituting all the values in above equation,
v = 1.28 m/s
Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.
Answers is F=7.84 N
Friction force resists the effect of horizontal force and trying to approch to a limiting force.
we have formula for limiting friction force between block and floor
F=Ц N
where N=mg
putting values we get answer.
Answer:
No, the farmer is not able to move the mule.
Explanation:
Mass =100 kg
Force=F=800 N
The coefficient between the mule and the ground=
Static friction force,f=
Normal force=N=mg
Static friction force,f=
Using 
F<f
Static friction force is greater than applied force.
Therefore , the farmer is not able to move the mule.
Refer to the diagram shown below.
Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.
The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.
The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s
The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°
Answer: 10.2 m/s at 21.5° downstream.
