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dem82 [27]
3 years ago
14

A computational model predicts the maximum potential energy a roller coaster car can have given its mass and its speed at the lo

west point of its journey. Its mass is 350 kg, and it speed at a height of 0 m is 25 m/s. What is the prediction for its maximum potential energy?
109,375 J

765,625 J

218,750 J

87,750 J
Physics
1 answer:
andreyandreev [35.5K]3 years ago
3 0

Answer:

The prediction for its maximum potential energy is 109,375 J

Explanation:

Given;

mass of the coaster car, m = 350 kg

speed of the coaster car at the lowest point, v = 25 m/s

The coaster car will have maximum kinetic energy at the lowest point and based on law of conservation of mechanical energy, the maximum kinetic energy of the coaster car at the lowest point will be equal to maximum potential energy at the highest point.

K.E_{max} = P.E_{max}

K.E_{max} = \frac{1}{2} mv^2\\\\K.E_{max} = \frac{1}{2} (350)(25)^2\\\\K.E_{max} =109,375 \ J

Thus, P.E_{max} = 109,375 \ J

Therefore, the prediction for its maximum potential energy is 109,375 J

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"Folders allow you to organize and maintain files" is the statement that best describes the role of folders in the operating system. The correct option in respect to the question given is option "D". Folders help a person to assimilate all the required files within one common space. It helps the person to find all documents related to a single subject within a single folder. If the folder was not present, then there was probality of missing an important file or loosing valuable time searching through a huge number of files. The folder can easily be moved from one drive to another without any hassle.
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3 years ago
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A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a
Zielflug [23.3K]

Answer:F_{v} =\mu_{k} mg

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to  start crate moving is F_{s} =\mu_{s} mg.

but once crate starts moving the force of friction is reduced  F_{v} =\mu_{k} mg.

Hence  to keep crate moving at constant velocity we have to reduce the  force pushing crate ie F_{v} =\mu_{k} mg.

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as  forces are balanced.

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4 0
4 years ago
a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the
erma4kov [3.2K]

The emerging velocity of the bullet is <u>71 m/s.</u>

The bullet of mass <em>m</em> moving with a velocity <em>u</em>  has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.

If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,

W=Fx

This is equal to the change in the bullet's kinetic energy.

W=Fx=\frac{1}{2} m(u^2-v^2)......(1)

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>

Assuming the same resistive forces to act on the bullet,

F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)

Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s

Thus the speed of the bullet is 71 m/s


3 0
3 years ago
Is the strange solid/liquid changing behavior of Oobleck the SAME as an ice cube melting?
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I don’t think so because an ice cube melting needs heat and relies on temp while ooblecks transition from solid to quickly depends on force and speed
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ANSWER ASAP
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The weight of the plane is  (mass) x (acceleration of gravity).
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