The magnitude of the magnetic force on a current-carrying wire held in a magnetic
1 answer:
Answer:
All of the above
Explanation:
The magnitude of the magnetic force on a current-carrying wire held in a magnetic is given by the equation
Where B = Strength of the magnetic field
I = The current carried by the wire
l = length of the wire in the magnetic field
θ = Angle between the wire and the magnetic field
Based on the relationship written above, the magnitude of the magnetic force on the current - carrying wire in the magnetic field depends on the strength of the magnetic field (B), length of the wire(l), current in the wire (I).
All the options are correct.
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Answer:
, the minus meaning west.
Explanation:
We know that linear momentum must be conserved, so it will be the same before () and after (
) the explosion. We will take the east direction as positive.
Before the explosion we have .
After the explosion we have pieces 1 and 2, so .
These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.
Since we know momentum must be conserved we have:
Which means (since we want and
):
So for our values we have:
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Answer:
The spring constant of the spring is 47.62 N/m
Explanation:
Given that,
Mass that is attached with the spring, m = 29 g = 0.029 kg
The spring makes 20 complete vibrations in 3.1 s. We need to find the spring constant of the spring. We know that the number of oscillations per unit time is called frequency of an object. So,
f = 6.45 Hz
The frequency of oscillator is given by :
k is the spring constant
k = 47.62 N/m
So, the spring constant of the spring is 47.62 N/m. Hence, this is the required solution.