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Kay [80]
3 years ago
14

What happens to a shopping cart if you get it rolling and then release it?

Physics
1 answer:
Anit [1.1K]3 years ago
7 0

Answer:

It will slowly come to a stop as long as it is moving on a flat surface

Explanation:

If the cart is moving on a flat surface, it will slowly come to a stop due to the friction forces acting on its wheels.

If the cart is on an inclined surface, it may gain acceleration enough to overcome any friction force, and thus continue with an accelerated motion.

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Using the videos and websites provided, or other online resources you find, and analyze the events of the Biosphere 2 story and
Andru [333]
Your answer has to be B
4 0
3 years ago
Read 2 more answers
The vertebral region is _________ to the scapular region.
Lisa [10]

Answer:

<em>The answer is medial!</em>

Explanation:

<em>The vertebral region is </em><u><em>medial</em></u><em> to the scapula.</em>

<em>Hope This Helps!</em>

<em>-</em><u><em>Justin:)</em></u>

7 0
1 year ago
A rolling ball with a mass of 1.01 kg crashes into a wall. The wall exerts a force of 2.40 Newtons and stops the ball. What was
klemol [59]

Answer:

Acceleration, a = 2.38m/s²

Explanation:

Given the following data;

Mass = 1.01 kg

Force = 2.4N

To find the acceleration;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

F = ma

Where;

  • F represents force.
  • m represents the mass of an object.
  • a represents acceleration.

Making acceleration (a) the subject, we have;

Acceleration, a = \frac{F}{m}

Substituting into the equation, we have;

Acceleration, a = \frac{2.4}{1.01}

<em>Acceleration, a = 2.38m/s²</em>

<em>Therefore, the magnitude of the acceleration of the ball at the time of the crash is 2.38 meters per seconds square. </em>

3 0
2 years ago
A ball is thrown upwards and returns to the same location. Compared with its initial speed its speed when it returns is about___
Strike441 [17]

Answer:

Explanation:

Let the ball is thrown upwards with speed u and reaches upto height h.

Use III equation of motion

v^{2}=u^{2}-2gh

here, final velocity v = 0

So, u = \sqrt{2gh}

Now as the ball reaches the maximum height, it starts falling freely. Let it strkes with the ground with velocity v'.

v'^{2}=u^{2}-2gh

here, u = 0

So, v' = \sqrt{2gh}

It means the velocity is same as the velocity of projection.

4 0
3 years ago
2.2 x 10^-5 um = _______ km
drek231 [11]
The answer is 2.2 × 10^-14<span> kilometers

hope this helps</span>
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3 years ago
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