The correct answer to the question is vertically downward i.e towards the centre of earth.
EXPLANATION:
As per the question, the box is pulled to the right.
Hence, the direction of the applied force is towards right.
We are asked to determine the direction of the gravitational force that acts on the body.
Before answering this question, first we gave to understand the gravitational force of earth.
Any body present on the surface of earth is attracted with the force of gravity of earth ( gravitational force ) towards its centre. It is equivalent to the weight of the body.
The force of gravity is always directed towards the centre of earth irrespective of the nature of applied force.
Hence, the direction of the gravitational force which acts on the box is vertically downward.
Answer:
The work done is 5136.88 J.
Explanation:
Given that,
n = 1.90 mol
Temperature = 296 K
If the initial volume is V then the final volume will be V/3.
We need to calculate the work done
Using formula of work done
Put the value into the formula
The Work done on the system.
Hence, The work done is 5136.88 J.
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Solution:
The angle between the sling and the load is 
So the tension in each sling can be calculated as
Where
M is the mass of the load
The Horizontal reaction on the sling will be inward.
After using the spreader, the new angle between sling and load is 
, the tension in the sling will be
= 
The tension will be same as before in the sling move away through the spreader at an angle more than 90 degree the horizontal force will act opposite and will be outward
Answer:
343/1500
Explanation:
Power: This can be defined as the product force and velocity. The S.I unit of power is Watt (w).
From the question,
P' = mg×v................. Equation 1
Where P' = power used to gain an altitude, m = mass of the engine, g = acceleration due to gravity of the engine, v = velocity of the engine.
Given: m = 700 kg, v = 2.5 m/s, g = 9.8 m/s²
Substitute into equation 1
P' = 700(2.5)(9.8)
P' = 17150 W.
If the full power generated by the engine = 75000 W
The fraction of the engine power used to make the climb = 17150/75000
= 343/1500
