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Thepotemich [5.8K]
3 years ago
14

. A force of 3.0 N acts through a distance of 12 m in

Physics
1 answer:
musickatia [10]3 years ago
6 0

Answer:

<h3>The answer is 36 J</h3>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 3 × 12

We have the final answer as

<h3>36 J</h3>

Hope this helps you

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1. A 25.35-g piece of iron absorbs 1562.75 Joules of heat energy, and its temperature
Lera25 [3.4K]

Answer:

Q = C M T        where C is the specific, M the mass, T the temperature change

Note 1 cal = 4.19 Joules

1562.75 J / (4.19 J/cal) = 378 cal

C = Q / (M * T) = 378 cal / (25.35 g * 155 deg C)

C = .096 cal / g deg C

8 0
1 year ago
In concave lenses a Distance object appears ​
JulsSmile [24]

Answer:

it appears to be farther away than it actually is, and therefore smaller then the object itself.

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3 years ago
You are part of the designing team for a new solar thermal power plant. The plant will use thermal energy storage to produce ele
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Explanation:

Below is an attachment containing the solution.

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3 years ago
A crowbar having the length of 1.75m is used to balance a load between the fulcrum and the load is 0.5m calculate MA
aliina [53]

Answer:

1.70

Explanation:

substract 1.75-0.5

3 0
3 years ago
A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me tempe
Simora [160]

Answer:

Final temperature is 295K

Explanation:

Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C×\frac{1mol}{63,546g}×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C×\frac{1mol}{18,02g}×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK×\frac{1mol}{63,546g}×25,0g×(X-363K) = -75,2J/molK×\frac{1mol}{18,02g}×100,0g× (X-293K)

9,64X J/K - 3499J = - 417X J/K + 122273J

426,64X J/K = 125772 J

<em>X = 295K</em>

<em></em>

Final temperature is 295K

I hope it helps!

6 0
3 years ago
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