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uysha [10]
3 years ago
7

If the roller coaster car in the above problem were moving at twice the speed, then what would be its new kinetic energy

Physics
1 answer:
Xelga [282]3 years ago
8 0

Answer:

it would move faster

Explanation:

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A measure of the amount of light received on Earth is a star's ____.
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Apparent magnitude because Apparent is kinda like when something is obvious. So that's how I think of it.

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A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that
julsineya [31]

Since angular speed is persistent, v/r is constant.
Where:

 v is tangential speed; and

 r is distance from axis.

 

Then equate v/r in both cases to get v in the second case.

Hence, speed = 2.2 x 2.1 / 1.4 meters

= 3.3 meters / seconds

 

Alternative solution would be:

w = 2.2 / 1.4 = 1.57

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2 years ago
1. 20.0 mL of a liquid has a mass of 58.2 g. What is the density of the liquid?
skelet666 [1.2K]

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2 years ago
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A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

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

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

6 0
2 years ago
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