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:
Showing results for Two point charge q, separated by 1.5cm have change value of +2.0 and -4.0AND/C respectively what is the magnitude of the Electric force midway between them?
Search instead for Two point charge q, seperated by 1.5cm have change value of +2.0 and -4.0N/C respectively what is the magnitude of the Electric force midway between them?
Answer:
is it 3?
Explanation:
Im taking a guess and just dividing 6 and 2
Answer
A. the work done on the refrigerant in each cycle is 105kJ
B the coefficient of performance of the refrigerator is 4.8
Explanation
Given data
Work done at high temperature T2 Qh=610kJ
Work done at low temperature T1 Ql=505kJ
We know that the net work done by the refrigerator is expressed as
Wnet= Qh-Ql
=610-505
=105kJ
Also we know that the coefficient of performance is expressed as
COP= Ql/Wnet
COP= 505/105
= 4.8
Is that a question? .
..............................
