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andrew-mc [135]
2 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]2 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]2 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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Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 19.0 ∘ west of north,
valina [46]

Answer:

=2.99\times10^9J

Explanation:

According to the question

net force F = 2.20×10^6 N

displacement S = 0.72\times10^3m

from figure , the horizontal forces are same in magnitude and opposite direction.

so , neglect these two forces.

we can take only vertical components of the force.

total force F' = F cos 19° + F cos 19°

= 2×F×cos 19°   ................. (1

therefore , total work is

W = F'S

= (2F cos19)×S

= (2)(2.20\times10^6 N)cos19° (0.720\times10^3 m)

=2.99\times10^9J

6 0
3 years ago
What is the science behind the making of pop-up books?
IRINA_888 [86]

Answer

The moveable parts of the pop up book are of often cut out by hand and are folded and glued by hand upon the printed pages. The cover is glued or sewn to the lining. Front and backs are often made up from board, which is just a heavier gauge paper than is used for the pages.

Explanation

Hope that this helps you and have a great day:)

6 0
3 years ago
Which planet moves the fastest
aksik [14]

Answer:

Explanation:

Mercury moves the fastest.

8 0
2 years ago
Read 2 more answers
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
2 years ago
A ship maneuvers to within 2500 m of an island's 1800 m high mountain peak and fires a projectile at an enemy ship 610 m on the
Ne4ueva [31]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

t=(0-(250sin75)^2)/-9.8 
<span>the distance one is (2500+610)- (250m/s*cos75)*t=Dh Dh=horizontal distance </span>

<span>the max height one is d=0.5*9.8*t^2 </span>
<span>d= max height subtract 1800-d</span>
3 0
2 years ago
Read 2 more answers
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