Answer:
<h3>The maximum possible torque is τ =
![4.8 \times 10^{-3}](https://tex.z-dn.net/?f=4.8%20%5Ctimes%2010%5E%7B-3%7D)
N .</h3>
Explanation:
Given :
No. of turns ![N = 52.5](https://tex.z-dn.net/?f=N%20%3D%2052.5)
Radius of circular coil
m
Magnetic field
T
Current
A
For maximum torque,
τ = ![BIAN \sin \theta](https://tex.z-dn.net/?f=BIAN%20%5Csin%20%5Ctheta)
Where
area of coil
=
For maximum possible torque
,
τ = ![52.5 \times 26.3 \times 10^{-3} \times 0.465 \times 75.39 \times 10^{-4}](https://tex.z-dn.net/?f=52.5%20%5Ctimes%2026.3%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%200.465%20%5Ctimes%2075.39%20%5Ctimes%2010%5E%7B-4%7D)
τ =
N
Answer:
Explanation:
As she danced, Cinderella's foot got sweaty, providing lubrication, and while running her fit simply came out. She was likely barefoot in the slipper, so this makes sense. The Fairy Godmother designed the shoe to come off at the critical moment.
Answer:
1.23 m/s^2
Explanation:
Knowing that mg= Weight force, plug in our values 55kg*g=67.9. Divide both sides by 55 and you will get the value of acceleration due to gravity (g) which = 1.23 m/s^2
<span>Fast-moving cold fronts force the warm air ahead of them to rise more quickly than slow movers. The clouds and rain are mostly along the front or ahead of it, and heavy rain or thunderstorms are common. A slow-moving front lifts the warm air more gradually; the rain is less intense, and the clouds form along and *behind* the surface cold front.</span><span>
</span>
The man is standing on the plank. Then he is pulling the rope towards himself. He pushes the plank forward with the legs so that he can pull the rope backwards.
The man is standing on the plank. Then he is pulling the rope towards himself. He pushes the plank forward with the legs so that he can pull the rope backwards. So the friction f acts in the forward direction for the plank. Also f acts in the opposite direction on the man. The tension T in the rope is 100 N (given). Let the friction force = f Newtons.
The man is standing on the plank. Then he is pulling the rope towards himself. He pushes the plank forward with the legs so that he can pull the rope backwards. So the friction f acts in the forward direction for the plank. Also f acts in the opposite direction on the man. The tension T in the rope is 100 N (given). Let the friction force = f Newtons.Let the common acceleration = a m/s^2
Man: <em>net force = T - f = m a = 50 a </em>
<em>net force = T - f = m a = 50 a </em>
<em>Plank: net force = T + f = m a = 100 a</em>
<em>net force = T - f = m a = 50 a </em>
<em>Plank: net force = T + f = m a = 100 a</em>
<em>As T = 100 N, a = 4/3 m/s^2 and f = 100/3 Newtons. </em>
[correct me if I'm wrong]:)