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
Hi There,
This is False.
Hope this helped!
Answer:
1. 75N
2. 67,983 J (=67.98 kJ)
Explanation:
1. Work = Force x Distance
we are given that Work = 1,500J and Distance = 20m
hence,
Work = Force x Distance
1,500 = Force x 20
Force = 1,500 ÷ 20 = 75N
2. Potential Energy, PE = mass x gravity x change in height
we are given that mass = 165 kg and change in height = 42m
assuming that gravity, g = 9.81 m/s²
Potential Energy, PE = mass x gravity x change in height
Potential Energy, PE = 165 x 9.81 x 42 = 67,983 J (=67.98 kJ)
Answer:
12
Explanation:
600/50 is 12 so the answer is twelve
Answer:0.27
Explanation:
Given
One worker Pushes with force 
other Pulls it with a rope of rope 
mass of crate 
both forces are horizontal and crate slides with a constant speed
Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it
where 
is the friction force
where 
is the coefficient of static friction
