Explanation:
<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>
Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass
as


Substituting (2) into (1), we get

where
, the frictional force on
Set this aside for now and let's look at the forces on 
<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>
Let the x-axis be (+) up along the inclined plane. We can write the forces on
as


From (5), we can solve for <em>N</em> as

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

Substituting (7) into (4) we get

Collecting similar terms together, we get

or
![a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)](https://tex.z-dn.net/?f=a%20%3D%20%5Cleft%5B%20%5Cdfrac%7Bm_B%5Csin30%20-%20%5Cmu_km_A%7D%7B%28m_A%20%2B%20m_B%29%7D%20%5Cright%5Dg%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%288%29)
Putting in the numbers, we find that
. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get
. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get 
Answer:
The magnetic field strength of an electromagnet is therefore determined by the ampere turns of the coil with the more turns of wire in the coil the greater will be the strength of the magnetic field.
Explanation:
It’s designed to protect an electrical circuit from damage caused by overcurrent, usually resulting from an overload or short circuit. Its basic function is to interrupt current flow after a fault is detected.
That’s really just the basic purpose.
Happy to help!
~Brooke❤️
Answer:

Explanation:
means initial angular velocity, which is 0 rev/min
means final angular velocity, which is 
t means time t= 3.20 s
one revolution is equivalent to 2πrad so the final angular velocity is:
= (2π/60) *2.513*10^{4} rad/s
= 2628.5 rad/s
so the angular acceleration, α will be:
α = 2628.5 rad/s / 3.20 s

so the rotational motion about a fixed axis is:
+ 2αΔTita where ΔTita is the angle in radians
so now find the ΔTita the subject of the formula
ΔTita = 

