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zysi [14]
2 years ago
9

two mass Ma=2kg and Mb=5kg on incline are connected together by string as shown below the coefficient of kinetic friction betwee

n each mass and it's incline is 0.3 if Ma move up and Mb move down determine. (1) the tension in the connecting string. (2)their acceleration​
Physics
1 answer:
aniked [119]2 years ago
5 0

Answer:

yourjejejrjrjrjr mother

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A manufacturer supplies plastic cups that are placed under the legs of the chair. The manufacturer's claim is that there will be no marks of chair due to pushing or pulling on the floor.

<h3>Who is a manufacturer?</h3>

A manufacturer is a person who converts his idea of developing and constructing any item using some set of machines.

A manufacturer supplies plastic cups that are placed under the legs of the chair.

When there is a relative motion between two objects, friction force causes the opposition to this motion. Between chair and floor, there will be high magnitude of force. But if we insert plastic cups in the legs of chair, then the friction between cup and floor will be less. This keeps the floor protected form any scratches.

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2 years ago
Which equation best describes the law of conservation of momentum?
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The law of conservation of momentum<span> states that for two objects colliding in an isolated system, the total </span>momentum<span> before and after the collision is equal. Momentum should be conserved. Hope this answers the question. Have a nice day.</span>
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3 years ago
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What cause earthquakes?
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Question; what causes earthquakes? 
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A locomotive approaches its next stop and accelerates at -0.12 m/s^2, coming to a complete stop in 30 seconds. This motion could
Masja [62]

Answer:

<em>Answer: positive velocity & negative acceleration</em>

Explanation:

<u>Accelerated Motion</u>

Both the velocity and acceleration are vectors because they have magnitude and direction. When the motion is restricted to one dimension, i.e. left-right or up-down, the direction is marked with the sign according to some preset reference.

The locomotive is moving at a certain speed with a (so far) unknown sign but the acceleration has a negative sign. Since the locomotive comes to a complete stop it means the velocity and the acceleration are of opposite signs.

Thus the velocity is positive.

Answer: positive velocity & negative acceleration

4 0
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:

W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]

W_{loss} = 6574.75\,J

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.

8 0
2 years ago
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