Answer:
4.8 turns would be made around the tube
Explanation:
You need the circumference of the tube since it's just a lifted circle
2×pi×3=18.85
90.42/18.85=4.79
And round that to 4.8
Answer:
76969.29 W
Explanation:
Applying,
P = F×v............. Equation 1
Where P = Power, F = force, v = velocity
But,
F = ma.......... Equation 2
Where m = mass, a = acceleration
Also,
a = (v-u)/t......... Equation 3
Given: u = 0 m/s ( from rest), v = 12.87 m/s, t = 3.47 s
Substitute these values into equation 3
a = (12.87-0)/3.47
a = 3.71 m/s²
Also Given: m = 1612 kg
Substitute into equation 2
F = 1612(3.71)
F = 5980.52 N.
Finally,
Substitute into equation 1
P = 5980.52×12.87
P = 76969.29 W
Answer:
The rate at which power is generated in the coil is 10.24 Watts
Explanation:
Given;
number of turns of the coil, N = 160
area of the coil, A = 0.2 m²
magnitude of the magnetic field, B = 0.4 T
time for field change = 2 s
resistance of the coil, R = 16 Ω
The induced emf in the coil is calculated as;
emf = dΦ/dt
where;
Φ is magnetic flux = BA
emf = N (BA/dt)
emf = 160 (0.4T x 0.2 m²)/dt
emf = 12.8 V/s
The rate power is generated in the coil is calculated as;
P = V²/ R
P = (12.8²) / 16
P = 10.24 Watts
Therefore, the rate at which power is generated in the coil is 10.24 Watts
I believe the answer is D. <span>The hypothesis is revised and another experiment is conducted.</span>
<span>Even though the Sun has a greater mass than Earth, the Moon orbits Earth because it's closer to the Earth than to the Sun. Because of this proximity between the Earth and the Moon, the Earth has a stronger gravitational pull than the Sun does. Furthermore, the Earth's mass is 81 times that of the Moon, and so at this proximity, it is more than able to overpower what pull the Sun exerts on the Moon.</span>
