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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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A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie
cluponka [151]

Answer:

a) v₃ = 19.54 km, b)  70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

         cos 45 = x₁ / V₁

         sin 45 = y₁ / V₁

         x₁ = v₁ cos 45

         y₁ = v₁ sin 45

         x₁ = 26 cos 45

         y₁ = 26 sin 45

         x₁ = 18.38 km

         y₁ = 18.38 km

Vector 2 moves 45 km north

        y₂ = 45 km

Unknown 3 vector

          x3 =?

          y3 =?

Vector Resulting 70 km north of the starting point

           R_y = 70 km

we make the sum on each axis

X axis

      Rₓ = x₁ + x₃

       x₃ = Rₓ -x₁

       x₃ = 0 - 18.38

       x₃ = -18.38 km

Y Axis

      R_y = y₁ + y₂ + y₃

       y₃ = R_y - y₁ -y₂

       y₃ = 70 -18.38 - 45

       y₃ = 6.62 km

the vector of the third leg of the journey is

         v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

         v₃ = √ (18.38² + 6.62²)

         v₃ = 19.54 km

to find the angle let's use trigonometry

           tan θ = y₃ / x₃

           θ = tan⁻¹ (y₃ / x₃)

           θ = tan⁻¹ (6.62 / (- 18.38))

           θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

          θ’= 180 -19.8

          θ’= 160.19º

I mean the address is

          θ’’ = 90-19.8

          θ = 70.2º

70.2º north-west

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