Answer:
0.243 m/s
Explanation:
From law of conservation of motion,
mu+m'u' = V(m+m')................. Equation 1
Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.
make V the subject of the equation
V = (mu+m'u')/(m+m')................. Equation 2
Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s
Substitute into equation 2
V = (260000×0.32+52500×(-0.14))/(260000+52500)
V = (83200-7350)/312500
V = 75850/312500
V = 0.243 m/s
The magnitude of the magnetic force per unit length on the top wire is
2×10⁻⁵ N/m
<h3>How can we calculate the magnitude of the magnetic force per unit length on the top wire ?</h3>
To calculate the magnitude of the magnetic force per unit length on the top wire, we are using the formula
F= 
Here we are given,
= magnetic permeability
= 4
×10⁻⁷ H m⁻¹
If= 12 A
d= distance from each wire to point.
=0.12m
Now we put the known values in the above equation, we get
F= 
Or, F = 
Or, F= 2×10⁻⁵ N/m.
From the above calculation, we can conclude that the magnitude of the magnetic force per unit length on the top wire is 2×10⁻⁵ N/m.
Learn more about magnetic force:
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I might be wrong but I'm pretty sure It's B
Answer:
Workdone = 1250Nm
Explanation:
<u>Given the following data;</u>
Force, F = 500N
Distance, d = 2.5m
Workdone is given by the formula;
Substituting into the equation, we have
Workdone = 1250Nm
Therefore, the actual work done by the worker is 1250 Newton-meter.