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zubka84 [21]
3 years ago
11

If you wanted to duplicate conditions in a laboratory that produce sedimentary rock, what would you need to do?

Physics
2 answers:
kifflom [539]3 years ago
7 0

Answer:

Apply high pressure and high temperature to the rock

Explanation:

Nookie1986 [14]3 years ago
7 0
Put higher temp on the rock, hope this helps
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the moderate temperatures of islands throughout the world has much to do with water's vast supply of internal energy. high evapo
Ad libitum [116K]

High specific heat of the water. Option (c)

What is Specific heat?

The amount of heat required to increase the temperature of one gram of a substance by one degree Celsius is referred to as the substance's specific heat. Typically, calories or joules are used per gram and degree Celsius when referring to the units of specific heat.

The moderate temperature of islands has much to do with the water's high specific heat. The typical off-water is more significant than this clear land or soil. Due to this fact, water absorbs and releases eat more slowly. In comparison to the land.

Hence, the water has high specific heat.

To learn more about specific heat the link is given below:

brainly.com/question/12982780?

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3 0
1 year ago
Please help question in photo
bonufazy [111]

Answer:

A.) the car

Explanation:

hope this helped <3

8 0
3 years ago
two masses are separated by 1 m. suppose the masses are moved so they are 2 m apart how will the gravitational force change
Verdich [7]
The gravitational force would get stronger because the farther the two masses are separated the more gravitational force will be used to pull them together the closer they are the less gravitational pull is used to pull them together
5 0
3 years ago
A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe
SVETLANKA909090 [29]

Answer:

T_s = 6.8 degree C

Explanation:

As per thermal radiation we know that rate is heat radiation is given as

\frac{dQ}{dt} = \sigma eA (T^4 - T_s^4)

here we know that

T = 34 degree C = 307 K

A = 1.38 m^2

e = 0.557

\sigma = 5.67 \times 10^{-8} W/m^2K^4

\frac{dQ}{dt} = 120 J/s

now we have

120 = (5.67 \times 10^{-8})(0.557)(1.38)(307^4 - T_s^4)

120 = (4.36 \times 10^{-8})(307^4 - T_s^4)

T_s = 279.8 K

T_s = 6.8 degree C

5 0
3 years ago
Explain your answer. What parts of your hypothesis were strong correct? What parts were weak?
Anastaziya [24]

This question is from quizlet.

So better check this question!

5 0
2 years ago
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