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2 years ago

How many water (in kg) can raise its temperature from 23°C to 100°C by adding 21500 Joules of energy?

Physics

1 answer:

igomit [66]2 years ago

4 0

Answer: Approximately 8.0g of water

Explanation:

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A block of mass 5 kg is descending with a constant velocity on an inclined plane at 30°. What is the value of the frictional for

sattari [20]

Answer:28.9N

Explanation:

F=UR R=mg R=5×10 R=50N

U=tan© U=Tan30 U=1/√(3)

F=UR F=1/√(3) ×50

F=28.9N

4 0

2 years ago

How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 C?

fomenos

Answer:The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.

Explanation:

7 0

2 years ago

Two hypothetical discoveries in Part A deal with moons that, like Earth's moon, are relatively large compared to their planets.

ser-zykov [4K]

Answer:

Explanation:

Solution:

- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.

- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible.

6 0

2 years ago

If it takes 50n of force to lift a 450n what is the ma of the machine

otez555 [7]

If the machine is 100% efficient, then its
Mechanical Advantage is (450/50) = 9 .

If the machine is less than 100% efficient,
then the MA is more than 9 .

7 0

3 years ago

A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

Fed [463]

<h2>Option A is the correct answer.</h2>

Explanation:

Acceleration due to gravity

$g=\frac{GM}{r^2}$

G = 6.67 × 10⁻¹¹ m² kg⁻¹ s⁻²

Let mass of earth be M and radius of earth be r.

We have

$g=\frac{GM}{r^2}$

Now

A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

Mass of hypothetical planet, M' = M/2

Radius of hypothetical planet, r' = 2r

Substituting

$g'=\frac{GM'}{r'^2}\\\\g'=\frac{G\times \frac{M}{2}}{(2r)^2}\\\\g'=\frac{\frac{GM}{r^2}}{8}\\\\g'=\frac{g}{8}$

Option A is the correct answer.

6 0

3 years ago