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4 years ago

A 4500 kg car accelerates from rest to 45.0

Physics

1 answer:

Llana [10]4 years ago

8 0

The car undergoes an acceleration a such that

(45.0 km/h)² - 0² = 2 a (90 m)

90 m = 0.09 km, so

(45.0 km/h)² - 0² = 2 a (0.09 km)

Solve for a :

a = (45.0 km/h)² / (2 (0.09 km)) = 11,250 km/h²

Ignoring friction, the net force acting on the car points in the direction of its movement (it's also pulled down by gravity, but the ground pushes back up). Newton's second law then says that the net force F is equal to the mass m times the acceleration a, so that

F = (4500 kg) (11,250 km/h²)

Recall that Newtons (N) are measured as

1 N = 1 kg • m/s²

so we should convert everything accordingly:

11,250 km/h² = (11,250 km/h²) (1000 m/km) (1/3600 h/s)² ≈ 0.868 m/s²

Then the force is

F = (4500 kg) (0.868 m/s²) = 3906.25 N ≈ 3900 N

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Complete Question

Complete Question is attached below

Answer:

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Explanation:

From the question we are told that:

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Giving

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$q=176N/C m^2*1.558*10^{-12}c$

$q=1.558*10^{-9}c$

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