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3 years ago

A car has an acceleration of -5 m/s^2. Describe the car’s motion

Physics

1 answer:

AlladinOne [14]3 years ago

7 0

Explanation:

because the acceleration is negative, this indicates a deceleration (or slowing down) . Hence we can say that:

The car is decelerating (slowing down), i.e its velocity is decreasing, at a constant rate of 5m/s².

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Determine the change in velocity of a car that starts at rest and has a final velocity of 20 miles per second North

Katen [24]

Answer: The change in velocity is 20mph

Explanation: The change in velocity is the difference between the final velocity and the initial velocity.

The initial velocity is 0 and the final velocity is 20mph.

Using the formula dV=Vf-Vi

dV=20-0

dV=20mph North

5 0

3 years ago

If an astronaut throws an object in space, the object’s speed will _____

BigorU [14]

The object's speed will not change.

In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
$\sum F=ma$
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.

7 0

3 years ago

A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of

Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force is $a = \frac{F}{m} = \frac{V^2}{r}$

Gravitational force $F= \frac{Gm m}{d^2}$

we know that

$v = \frac{2\pi R}{T}$

solving for

$R = \frac{vT}{2\pi}$

$F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}$

$F = G\times \frac{\pi m}{(vT)^2}$

$a = \frac{v^2}{\frac{vT}{2\pi}}$

$a = \frac{2\pi v}{T}$

we know that

f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

solving for m

$m = \frac{2Tv^3}{\pi G}$

$m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}$

$m = 2.23 \times 10^{-32} kg$

5 0

3 years ago

What is the frictional force between a box and the floor it is being pulled across if, the kinetic coefficient of friction is 0.

Artyom0805 [142]

If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that

<em>n</em> = 38 N

The friction force is proportional to the normal force by a factor of 0.27, so that

<em>f</em> = 0.27 (38 N) ≈ 10.3 N

and so the answer is D.

8 0

3 years ago

Please help me with this question...

vredina [299]

The answer is B (The second one). I'm not sure though.

3 0

3 years ago