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3 years ago

A car is stationary. It accelerates at 0.8 ms^2

Physics

1 answer:

lara31 [8.8K]3 years ago

8 0

Answer:

the car’s final displacement is 60 m

Explanation:

Given;

initail velocity of the car, u = 0

acceleration of the car, a = 0.8 m/s²

time of motion, t = 10 s

The first displacement of the car:

$x_1 = ut + \frac{1}{2} at^2\\\\x_1 = 0 + \frac{1}{2} (0.8)(10)^2\\\\x_1 = 40 \ m$

The second displacement of the car;

acceleration, a = 0.4 m/s², time of motion, t = 10 s

$x_2 = ut + \frac{1}{2} at^2\\\\x_2 = 0 + \frac{1}{2} (0.4)(10)^2\\\\x_2 = 20 \ m$

The final displacement of the car;

x = x₁ + x₂

x = 40 m + 20 m

x = 60 m

Therefore, the car’s final displacement is 60 m

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Read 2 more answers

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djyliett [7]

Answer:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

$F = G \frac{m_{Earth} m_{Spaceship}}{r^2}$

Where G is a constant $G = 6.674 x10^{-11} m^2/ (ks s^2)$

$m_{Earth}$ represent the mass for the earth

$m_{spaceship}$ represent the mass for the spaceship

$r$ represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.