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Answer:
The appropriate solution is "61.37 s".
Explanation:
The given values are:
Boat moves,
= 10 m/s
Water flowing,
= 1.50 m/s
Displacement,
d = 300 m
Now,
The boat is travelling,
= 
= 
Travelling such distance for 300 m will be:
⇒ 

On putting the values, we get


Throughout the opposite direction, when the boat seems to be travelling then,
= 
= 
Travelling such distance for 300 m will be:
⇒ 

On putting the values, we get


hence,
The time taken by the boat will be:
= 
= 
Answer:
External force on him will be 112 N
Explanation:
We have given the mass of the sprinter m =70 kg
Acceleration of the sprinter 
We have to find the net external force
According to second law of motion force = mass ×acceleration
Force is dependent on the mass and acceleration
So 
So external force will be 112 N
Answer:
The magnitude of the applied torque is 
(e) is correct option.
Explanation:
Given that,
Mass of object = 3 kg
Radius of gyration = 0.2 m
Angular acceleration = 0.5 rad/s²
We need to calculate the applied torque
Using formula of torque

Here, I = mk²

Put the value into the formula



Hence, The magnitude of the applied torque is 