Answer:
a
Generally from third equation of motion we have that
Here v is the final speed of the car
u is the initial speed of the car which is zero
is the initial position of the car which is certain height H
is the final position of the car which is zero meters (i.e the ground)
a is the acceleration due to gravity which is g
So
=>
b
Explanation:
Generally from third equation of motion we have that
Here v is the final speed of the car
u is the initial speed of the car which is zero
is the initial position of the car which is certain height H
is the final position of the car which is zero meters (i.e the ground)
a is the acceleration due to gravity which is g
So
=>
When we have that
=>
=>
Answer:
E = 2.17 x 10⁻² V/m
Explanation:
First we will find out the current density by using the formula:
J = I/A
where.
J = Current Density = ?
I = Current = 5.5 A
A = Cross-Sectional Area = πr² = π(1.5 x 10⁻³ m)² = 7.068 x 10⁻⁶ m²
Therefore,
J = 5.5 A/7.068 x 10⁻⁶ m²
J = 0.778 x 10⁶ A/m²
Now, we calculate the magnitude of applied field:
E = ρJ
where,
E = Magnitude of applied field = ?
ρ = resistivity of Aluminum = 2.8 x 10⁻⁸ Ω.m
Therefore,
E = (2.8 x 10⁻⁸ Ω.m)(0.778 x 10⁶ A/m²)
<u>E = 2.17 x 10⁻² V/m</u>