Answer:
The maximum torque on the loop is 395.80 N.m.
Explanation:
Given;
number of turns of the wire, N = 150 turns
length of the square loop, L = 18.0 cm = 0.18 m
current in the wire, I = 50.9 A
Magnetic field, B = 1.6 T
Maximum torque on the loop is given by;
τ = NIAB
τ = (150)(50.9)(0.18²)(1.6)
τ = 395.80 N.m
Therefore, the maximum torque on the loop is 395.80 N.m.
The nervous system and the circulatory system
Answer:
0.50m/s
Explanation:
Average velocity is the change in displacement of a body with respect to time.
Velocity = ∆S/∆t
∆S = 100m - 70m
∆S = 30m
∆t = 2min - 1 min
∆t = 1min = 60secs
Substitute the given parameters into the formula for velocity
Velocity = 30m/60s
Velocity = 1/2 m/s
Average Velocity = 0.5m/s
The car on the top of the arc;
m = 1,400 kg, v = 15 m/s, r = 40 m;
The normal force:
F n = m g - m v ² / r =
= 1,400 kg · 9.8 m / s² - (1,400 kg · ( 15 m/s )² : 40 m ) =
= 13,720 N - 7,875 N = 5,845 N
Answer:
B ) 5,800 N
B you will see the objective outside the vehicle not moving will’l you are moveing inside the vehicle
