What is true of the moon's orbital and rotational periods?
1 answer:
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Answer:
the magnitude of the force that the wire will experience = 1.8 N
Explanation:
The force on a current carrying wire placed in a magnetic field is :
F = Idl × B
where:
I = current flowing through the wire
dl = length of the wire
B = magnetic field
We can equally say that :
where : sin θ is the angle at which the orientation from the magnetic field to the wire occurs = 30°
Then;
Given that:
L = 20 cm = 0.2 m
I = 6 A
B = 3 T
θ = 30°
Then:
F = 3 × 6 × 0.2 sin 30°
F = 1.8 N
Therefore, the magnitude of the force that the wire will experience = 1.8 N
Answer:
11060M Joules, where M is the mass of the diver in kg
Explanation:
Mass of the skydiver missing, we're assuming it's M.
It's total energy is the sum of the contribution of his kinetic energy (K)- since he's moving at 50 m/s, and it's potential energy (U), since he's subject to earth gravity.
Energy is the sum of the two, so
Answer:
a) KE = 888.26J
b) N = 294.5 turns
Explanation:
For the kinetic energy:
The inertia is:
So, the kinetic energy will be:
Now, friction force is:
Ff = μ*N = 0.80*5N = 4N
The energy balance would be:
Kf - Ko = Wf where Kf=0; Ko = 888.26J; and Wf is the work done by friction force.
Wf = -Ff*d = -Ff*N*2*π*R where N is the amount of turns it gives.
Replacing these values into the energy balance:
0-888.26=-4*N*2*π*0.12
-888.26=-0.96*π*N
N=294.5 turns