Answer:
The magnitude of the induced Emf is 
Explanation:
The width of the truck is given as 79inch but we need to convert to meter for consistency, then
The width= 79inch × 0.0254=2.0066 metres.
Now we can calculate the induced Emf using expresion below;
Then the 
Where B= magnetic field component
L= width
V= velocity
=(40*10^-6) × (42) × (2.0066)
=0.003371V
Therefore, the magnitude emf that is induced between the driver and passenger sides of the truck is 0.003371V
We know that a charge moving in a magnetic field is subject to the force:
F = q · v · B
But we also know that:
F = m · a
Therefore, it must be:
m · a = <span>q · v · B
And solving for a:
</span>a = <span>q · v · B / m
Recall that for a proton:
q = 1.6</span>×10⁻¹⁹ C
m = 1.673×10⁻²⁷ kg
Now, you can find:
a = 1.6×10⁻¹⁹ · 7.0 · 1.7 / <span>1.673×10⁻²⁷
= 1.14</span>×10⁹ m/s²
Hence, the acceleration of the proton is 1.14<span>×10⁹ m/s²</span>.
For example, an internal combustion engine converts<span> the </span>potential<span> chemical </span>energy <span>in gasoline and oxygen into </span>thermal energy<span> which, by causing pressure and performing work on the pistons, is </span>transformed<span> into the mechanical </span>energy<span> that accelerates the vehicle (increasing its kinetic </span>energy<span>).</span><span />
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Answer:we can calculate the answer by following steps;
Explanation: