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andrew-mc [135]
3 years ago
9

A thermal reservoir can be characterized as a thermal body that is large enough that when energy is dumped into it or taken out

of it, the __ of the reservoir does not vary considerably. The answer is an 11 letter word that begins with t and ends with e.
Physics
2 answers:
Ganezh [65]3 years ago
8 0

It should be noted that thermal reservoir is been characterized as a thermal body which is large enough and the removal of energy or dumping of it into it , the temperature of the reservoir does not vary considerably.

The thermal reservoir helps in the storage of energy, and it allows the removal as well as supplying of energy into it , however the temperature difference is not usually much.

Therefore, Temperature is the correct term.

learn more about thermal reservoir at;

brainly.com/question/13154260

hammer [34]3 years ago
7 0

A thermal reservoir can be characterized as a thermal body that is large enough that when energy is dumped into it or taken out of it, the temperature of the reservoir does not vary considerably.

<h3>What is a thermal reservoir?</h3>

A thermal reservoir is as described, a body large enough to have a very high heat capacity. This heat capacity refers to the amount of energy needed to raise the temperature by one degree.

Therefore, we can confirm that bodies with a large enough heat capacity will be considered thermal reservoirs. This is due to the fact that when energy is dumped into it or taken out of it, the temperature of the reservoir <u>does not vary considerably</u>.

To learn more about heat capacity visit:

brainly.com/question/1453843?referrer=searchResults

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Each magnet has a north pole and a south pole. We know that, from having played with bar magnets in our childhood, that a magnet's north pole will repel another magnet's north pole and attract its south pole.

From this diagram it is easy to see that the two lower bar magnets not only repel each other, but they are quite attracted to each other since their north and south poles are close together.

Therefore the region between the lower two magnets has the least force of repulsion.

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Are any rocks that form from another rock as a result of change in blank or blank
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4 years ago
Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m
mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0
3 years ago
A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the
Shalnov [3]

Heat required to raise the temperature of a given system is

Q = ms\Delta T

here we know that

m = mass

s = specific heat capacity

\Delta T = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2

Q = 5 * 1700 * 25 + 2 * 900 * 25

Q = 212500 + 45000

Q = 257500 J

So heat required to raise the temperature of the system is 257500 J

4 0
3 years ago
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