From the calculations, the power expended is 43650 W.
<h3>What is the power expended?</h3>
Now we can find the acceleration from;
v = u + at
u = 0 m/s
v = 95 km/h or 26.4 m/s
t = 6.8 s
a = ?
Now
v = at
a = v/t
a = 26.4 m/s/ 6.8 s
a = 3.88 m/s^2
Force = ma = 850-kg * 3.88 m/s^2 = 3298 N
The distance covered is obtained from;
v^2 = u^2 + 2as
v^2 = 2as
s = v^2/2a
s = (26.4)^2/2 * 3.88
s = 696.96/7.76
s = 90 m
Now;
Work = Fs
Work = 3298 N * 90 m = 296820 J
Power = 296820 J/ 6.8 s
= 43650 W
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The magnitude is 13.12 mV.
The steps are attached below.
<h3>How do you find the magnitude of an induced emf?</h3>
The standard SI unit of the magnetic field is the tesla (T). As an end result, we can use these equations and the equation for an induced emf due to changes in magnetic flux, ϵ=−NΔϕΔt ϵ = − N Δ ϕ Δ t, to calculate the importance of a precipitated emf in a solenoid.
The magnitude of the precipitated contemporary depends on the rate of trade of magnetic flux or the fee of reducing the magnetic area strains.
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Answer : The force will be 4501.9
We can see that, two forces acting on the dummy in two different direction.
We know that, here two forces are given in perpendicular direction with each other.
We know the force is the vector addition law so, we will use the Pythagoras theorem for the resultant of the vectors
Now, the net force is

Two forces is given,


Now, the net force is



Hence, the force will 4501.9 N