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satela [25.4K]
3 years ago
13

Blue bumper car, traveling east with a force of 100 N, runs into a yellow bumper car traveling west with a force of 300 N. Becau

se the net force will be -200 N. one of the cars will be pushed backward. Which one?
Physics
1 answer:
Pani-rosa [81]3 years ago
4 0

Answer:

the blue bumper car

Explanation:

because the yellow bumper car is coming in with much more force

and according to Newton, actions and reactions are equal and opposite, so the yellow bumper car will five out 300N while the blue car will give out only 100N

then it almost because a fight between 300N vs 100N

and obviously, the 300N will win and the 100N will be pushed back

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Two spherical shells have a common center. A -1.50 × 10-6 C charge is spread uniformly over the inner shell, which has a radius
Murrr4er [49]

Answer:

a) At 0.20 m, the magnitude of the field is 675.0 kV

The direction of the field is acting outwards from the center of the charged spheres

b) At 0.10 m, the magnitude of the field is 135 kV

The direction is acting outwards from the center of the charged spheres

c) At 0.025 m

The magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres

Explanation:

The charged spherical shell parameters are;

The charge on the inner sphere, q₁ = -1.50 × 10⁻⁶ C

The radius of the inner shell, R₁ = 0.050 m

The charge on the outer sphere, q₂ = +4.50 × 10⁻⁶ C

The radius of the outer shell, R₂ = 0.15 m

Let 'r', represent the distance at which the electric field is measured, the following relationships can be obtained;

When r < R₁ < R₂,

V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )

When R₁ < r < R₂,

V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )

When R₁ < R₂ < r,

V = k \cdot \left( \dfrac{q_1  + q_2}{r^2}  \right )

a) When r = 0.20 m, we have;

R₁ < R₂ < r, therefore

V = k \cdot \left( \dfrac{q_1  + q_2}{r^2}  \right )

By plugging in the values, we get;

V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} + 4.50\times 10^{-6} }{0.20^2}  \right ) = 675.0 \ kV

Therefore, the magnitude of the field, V = 675.0 kV

The direction of the field is outwards

b) When r = 0.10 m, we have;

When R₁ < r < R₂, therefore;

V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )

By plugging in the values, we get;

V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6}  }{0.10}  + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = 135 \ kV

Therefore, the magnitude of the field, V = 135 kV

The direction of the field is outwards from the center

c) When r = 0.025 m, we have;

When r < R₁ < R₂, therefore;

V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )

By plugging in the values, we get;

V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6}  }{0.05}  + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = -270 \ kV

Therefore, the magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres.

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