Answer:
T= 8.061N*m
Explanation:
The first thing to do is assume that the force is tangential to the square, so the torque is calculated as:
T = Fr
where F is the force, r the radius.
if we need the maximum torque we need the maximum radius, it means tha the radius is going to be the edge of the square.
Then, r is the distance between the edge and the center, so using the pythagorean theorem, r i equal to:
r = 
r = 0.5374m
Finally, replacing the value of r and F, we get that the maximun torque is:
T = 15N(0.5374m)
T= 8.061N*m
Answer:
The work done on the object by the force in the 5.60 s interval is 40.93 J.
Explanation:
Given that,
Force 
Mass of object = 2.00 kg
Initial position 
Final position 
Time = 4.00 sec
We need to calculate the work done on the object by the force in the 5.60 s interval.
Using formula of work done
Put the value into the formula
Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.
Since V is 5 times larger than C
A) 0V = 25C
25C to 0C = -25
-25 / 5 = -5
so 0C = -5 V
B)
20V x 5 = 100 C
Answer:
24.3KW
Explanation:
A)The kinetic energy is changing, the potential energy is changing and the chemical energy in form of fuel powering the engine also is changing
The kinetic energy is increasing as the body gain speed, the potential energy also increases as the body gain height against gravity and the chemical energy in form of fuel decreases as the body burn the fuel to create a lifting force
B) The workdone by the lifting force = the change in kinetic energy + the change in potential energy
C)The time taken in seconds to do the work is the variable needed
D) average power generated by the lifting force = (change in kinetic energy + change in potential energy) / time taken in seconds
Average power = 1/2 * m(mass) (Vf-Vi)^2 + mg(hf-hi) /t where vf is final speed and vi is initial speed at rest = 0, similarly, hf = final height and hi = initial height.
Average power = 1/2*810*7^2 + 810*9.81*8.2/3.5s
Average power = (19845+65158.02)/3.5 = 24286.577 approx 24.3kW
