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german
3 years ago
7

A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?

Physics
1 answer:
trasher [3.6K]3 years ago
3 0

Answer:

i dont know

Explanation:

Sorry and good luck

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A 70kg man runs at a pace of 4 m/s and a 50g meteor travels at 2 km/s. Which has the most kinetic energy?.
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Answer:

the 70kg man

Explanation:

because he has more weight and is moving faster

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A 200.0 kg piano is elevated by a crane to a height of 10.0 meters above a sidewalk. If the rope holding the piano breaks, what
Nitella [24]
All that business about the crane and the rope and the falling
is only there to confuse us.

The piano ended up 5 meters above the ground.

           Potential energy = (mass) (gravity) (height)

                                   = (200 kg) (9.81 m/s²) (5 m)

                                   = (200 · 9.81 · 5)  (kg-m²/s²)

                                   =   9,810 joules   . 
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Energy can be transformed from one form to another. The diagram shows one such process.
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8 0
3 years ago
Read 2 more answers
Plz answer will give Brainliest answer to whoever does.
Jobisdone [24]

Answer:

A foot kicking ball.

An iPhone being charged

Explanation:

  • A foot kicking a ball is example of transportation of Kinetic energy

\\ \sf\longmapsto K.E=\dfrac{1}{2}mv^2

  • An iPhone being charged means the electronic energy is converted into machinery energy
6 0
3 years ago
A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a
frosja888 [35]

Answer:

a) Yes

b) 7 rad/s

c) 0.01034 J

Explanation:

a)

Yes the angular momentum of the block is conserved since the net torque on the block is zero.

b)

m = mass of the block = 0.0250 kg

w₀ = initial angular speed before puling the cord = 1.75 rad/s

r₀ = initial radius before puling the cord = 0.3 m

w = final angular speed after puling the cord = ?

r = final  radius after puling the cord = 0.15 m

Using conservation of angular momentum

m r₀² w₀ = m r² w

r₀² w₀ = r² w

(0.3)² (1.75) = (0.15)² w

w = 7 rad/s

c)

Change in kinetic energy is given as

ΔKE = (0.5) (m r² w² - m r₀² w₀²)

ΔKE = (0.5) ((0.025) (0.15)² (7)² - (0.025) (0.3)² (1.75)²)

ΔKE = 0.01034 J

6 0
3 years ago
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