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horrorfan [7]
3 years ago
13

What is the length of an aluminum rod at 65°C if its length at 15°C is 1.2 meters?

Physics
2 answers:
lana66690 [7]3 years ago
3 0

<em><u>Answer:</u></em>

The new length is 1.20138 m

<em><u>Explanation:</u></em>

<u>1- We get the increase in length:</u>

The old length = 1.2 m

The coefficient of expansion of of aluminum is 23 * 10⁻⁶ /K.

The old temperature is 15°C.

The new temperature is 65°C

The increase in length= old length * coefficient of expansion *change in temperature

The increase in length = 1.2 * 23 * 10⁻⁶ * (65-15)

The increase in length = 0.00138 m

<u>2- getting the new length:</u>

new length = old length + increase in length

new length = 1.2 + 0.00138

new length = 1.20138 m

Hope this helps :)

Radda [10]3 years ago
3 0
<span>1.201386 meters at 65 degrees C.      
 Will be your answer!

Hope I Helped!</span>
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to move a resting box of 100 Newton on the ground with kinetic friction coeficient of 0,250 is applied a force of 60 N horizonta
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Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
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6 0
3 years ago
A brick in the shape of a cube with sides 10 cm has a density of 2500 kg/m^3. What is its weight? a.) 250 N
Arisa [49]

Answer:

c.) 25 N

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being l the side of the cube, which in this case is 10 cm or 0,1 m. Now we find the mass of the object, knowing the density and the Volume of the cube:

m=V*d\\m=(0,001 \:\:m^{3})(2500 \: \: \frac{Kg}{m^{3}})=2,5 \:\: Kg

We find the weight by multiplying the mass of the object with the gravity constant.

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8 0
3 years ago
A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te
Jet001 [13]

Answer: 1026s, 17.1m

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Given

COP of heat pump = 3.15

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Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

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Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

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3 years ago
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