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3 years ago

What is the kinetic energy of a 30 kg falling object when the object reaches a velocity of 20 m/s?

Physics

1 answer:

Oksi-84 [34.3K]3 years ago

6 0

Ek = 1/2mv^2 Ek = 1/2 * 30kg * 20^2 m/s Ek = 6000J Hope this Helps!

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Two identical blocks, A and B, are on a horizontal surface, as shown above. There is negligible friction between the surface and

e-lub [12.9K]

Answer:

the speed of the center of mass stays the same

Explanation:

In a system with no energy loss, momentum is conserved if the mass remains constant. The system described has no change in mass, and energy loss is considered negligible. Hence the product of the total mass and the velocity of its center will be a constant. The center of mass stays the same speed.

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3 years ago

Describe the relationship between a compass and a magnet. Use C.E.R while answering. Your CER should have a one-sentence claim,

Allushta [10]

Answer:

Magnets interfere with compasses because compasses use the magnetic field to locate NESW (north east south west)

Explanation:

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3 years ago

Read 2 more answers

Help please help! Need help

Alborosie

Answer:

M. Magnetism is a property of individual atoms.

Explanation:

when a magnet is broken into pieces the new pieces behave like the original magnet this observation shows that magnetism is the property of individual atoms.

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2 years ago

When sedimentary rock is exposed to heat and pressure, what does it change into?

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Metamorphic rock this possess often occurs in the mantle

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3 years ago

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and

nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}$

k = 1.4

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K$

Work done is given as;

$W = \frac{1}{2} *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

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3 years ago