Answer:
The magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
Explanation:
Given;
radius of the wire, r = 0.45 m
current on the loop, I = 2.4 A
angle of inclination, θ = 36⁰
torque on the coil, τ = 1.5 N.m
The torque on the coil is given by;
τ = NIBAsinθ
where;
B is the magnetic field
Area of the loop is given by;
A = πr² = π(0.45)² = 0.636 m
τ = NIBAsinθ
1.5 = (1 x 2.4 x 0.636 x sin36)B
1.5 = 0.8972B
B = 1.5 / 0.8972
B = 1.67 T
Therefore, the magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
The force exerted by a pressure of any gas over a surface its given by the formula P=F/S (where P is pressure, F force and S surface).
We can multiply both sides of the formula by S to obtain the force.
P*S=(F*S)/S
P*S=F
Solve for P=1.80*10^5 Pa and S=4.10*10^-4 m^2 ([Pa] =[N/m^s])
(1.80*10^5 N/m^s) * (4.10*10^-4 m^2) =F
73.8 N =F
Answer:
<h2>1116.9 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 438 × 2.55
We have the final answer as
<h3>1116.9 N</h3>
Hope this helps you
