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Aloiza [94]
2 years ago
15

A six pack of your favorite beverage (12 fl. oz. cans, assumed to have the same characteristics as water) is placed in a cooler

at 70° F. How much heat must be removed (BTU) to cool the beverages to 34°F?
Physics
1 answer:
nasty-shy [4]2 years ago
4 0

Answer:

Q = 169 BTU

Explanation:

As we know that volume is given as

V = 12 Fl oz

so it is given in liter as

V = 12 fl oz = 0.355 Ltr

now we have six pack of such volume

so total volume is given as

V = 6\times 0.355 Ltr

V = 2.13 ltr

so its mass is given as

m = 2.13 kg

now the change in temperature is given as

\Delta T = 70 - 34 = 36 ^oF

\Delta T = 20^oC

now the heat given to the liquid is given as

Q = ms\Delta T

Q = 2.13(4186)(20)

Q = 1.78 \times 10^5 J

Q = 169 BTU

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I have no idea of how to approach this problem
nataly862011 [7]

Answer:

p=1

Explanation:

Well me know that v=m/s

and that a=m/s^2

so

(m/s)^{2} = (m/s^2)(x^p)\\\\(m^2/s^2)/(m/s^2)=x^p\\\\m^2s^2/ms^2=x^p\\\\(m^2/m)(s^2/s^2)=x^p\\\\m=x^p\\\\p=1

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the gasoline in a car does 40000 j of work on a car and generates a constant force of 20 n. how far did the car go?
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     40,000 J  =  (20 N) x (distance)

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(20 N is not a huge force when it's being used to move a car.
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