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3 years ago

When can a car have no acceleration but still be moving

Physics

2 answers:

gregori [183]3 years ago

5 0

Answer:

A. A car can still be moving without acceleration when it has some velocity.

B. A car can come to a stop when moving itself when it has some de acceleration.

C.A ca can start at 0 m/s and speed up to 15 m/s when it has some acceleration.

Explanation:

A. since the car has no acceleration the velocity of the car would help it moving.

B. when the car is moving with some velocity but coming to a stop this is possible of and only if it has some de acceleration

C. since the speed of the car increases from 0 to 15 m/s acceleration is involved. If the speed was constant there would be no acceleration involved.

mylen [45]3 years ago

3 0

Answer:

When the velocity of the car is constant.

Explanation:

A car moving with no acceleration possesses a uniform velocity i.e the car velocity is not changing. This means that the car is covering a distance at a particular time interval "on a straight line". The velocity of the car is constant at that point that it has no acceleration because for a body to accelerates, the velocity of the body must be changing i.e the final velocity and the initial velocity must be different.

If the initial and final velocity of a body is the same, this means that the body is moving with a constant velocity and as such not accelerating.

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4 years ago

What is the force of a object that has a mass of 7 kg and an acceleration of 6 m/s/s

UkoKoshka [18]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

mass = 7 kg

acceleration = 6 m/s²

We have

force = 7 × 6 = 42

We have the final answer as

Hope this helps you

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3 years ago

Which compound has ionic bonding? A. Cl2 B. HF C. CaO D. NO2

GrogVix [38]

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3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is: