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Kamila [148]
3 years ago
9

The mass of an object with 500 J of kinetic energy moving with a velocity of 5 m/s is/

Physics
1 answer:
antiseptic1488 [7]3 years ago
7 0

Answer:

495

Explanation:

subtract 500 and 5

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atroni [7]
Do it in your own words
5 0
3 years ago
WILL MARK BRAINLIEST<br> If it takes you 70s to run the track, what is your velocity?
tiny-mole [99]

Answer:

10

Explanation:

But you have to be fat though lol

5 0
2 years ago
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
Two difference between thrust and upthrust .
velikii [3]

Upthrust is the upward force exerted by fluids on the surface of an object immersed in fluids. Thrust:- It is the force acting perpendicular to the surface. Upthrust:- It is the upward force exerted by the fluid on the surface of an object immersed in liquid.

I hope it's help you

6 0
3 years ago
Everyone open this please!
notsponge [240]

Answer:

yeah i knowwwwwwwwwwwwww

Explanation:

7 0
3 years ago
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