Answer:
Torque; τ = 4.712 × 10^(-3) J
Magnetic moment; M = 0.0248 J/T
Explanation:
Torque is gotten from the formula;
τ = BIA
Where;
B is magnetic field
I is current
A is area
We are given;
B = 0.19T
I = 6.2A
Rectangle dimensions = 5cm by 8cm = 0.05m by 0.08m
Thus;
Area; A = 0.05m × 0.08m = 0.004 m²
Thus;
τ = 0.19 × 6.2 × 0.004
τ = 4.712 × 10^(-3) J
Formula for the magnetic moment is given by;
M = IA
M = 6.2 × 0.004
M = 0.0248 J/T
Answer:
F = G M m / R^ force between m and earth where R >= radius of earth
a = F / m = G M / R^2 acceleration at radius R
am / ae = (Re / Rm)^2 acceleration of meteor to that of earth
am / ar = (4 / 9.8) = (Re / Rm)^2
Rm = (9.8 / 4)^1/2 Re
Rm = 1.56 Re = 1.56 * 6.4E106 m = 10E6
10E6 - 6.4E6 = 3.6E6 m above surface of earth
(You only need mass of earth if you are calculating a at the surface of the earth)
Calculate g = G M / R^2 = 6.67E-11 * 6E24 / 6.4E6)^2 = 9.77 m/s^2
This is close to the value we used - 9.8 m/s^2
Answer:
The answer is frictional force !
Newton's second law on motion states that a<span>cceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).
So to put it in super simple terms, p</span><span>ushing or pulling an object produces acceleration, a change in the speed of motion.</span>
I think this is correct, but I am not entirely certain.
Find the force constant of the spring:
F = - KX
(0 - 62.4) = -K(0.172m)
-362.791 = -K
362.791 N/m = K
Find the work done in stretching the spring:
W = (1/2)KX
W = (1/2)(362.791)(0.172m)
W = 31.2 J