Answer:
The magnitude of the torque on the loop due to the magnetic field is
.
Explanation:
Given that,
Diameter = 10 cm
Current = 0.20 A
Magnetic field = 0.30 T
Unit vector
We need to calculate the torque on the loop
Using formula of torque

Where, N = number of turns
A = area
I = current
B = magnetic field
Put the value into the formula


Hence, The magnitude of the torque on the loop due to the magnetic field is
.
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a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N
b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>
For this use the formula:
d = Vo * t - (at^2) / 2
Clearing t:
t = d/(v + 0.5*a)
Replacing:
t = 5 m / (7.2 m/s + 0.5 * (-1.1 m/s²)
Resolving:
t = 5 m / (7.2 m/s + (-0.55 m/s²)
t = 5 m / 6.65 m/s
t = 0.75 s
Result:
The time will be <u>0.75 seconds.</u>
Since the law of gravitation is an inverse square law if you
quadruple the radius the f will drop by a factor of 16 SO the object would
weigh 200/16 = 12.5N
In other words, as the distance, or radius, quadruples the
weight becomes 1/16 of the original weight. Just plug in 4 for r and when you
square it you get 16. The numerator is 1 so that is how the weight becomes
1/16.