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shepuryov [24]
2 years ago
11

Part a a calorimeter consists of an aluminum cup inside of an insulated container. The cup is weighed on a top-loading balance a

nd is found to have a mass of 31. 91 g. A reaction is conducted in the calorimeter, raising the temperature from 21. 2 c to 26. 1 c what is the change in heat q for the aluminum cup in units of j? aluminum has a specific heat of 0. 903 j 1 ? 1 write your answer to the correct number of significant figures.
Physics
1 answer:
vovikov84 [41]2 years ago
5 0

The change in heat q for the aluminum cup is 141 J.  

<h3>What is Specific heat?</h3>

A substance's heat capacity is the amount of heat needed to increase its overall temperature by one degree. The heat capacity is known as the specific heat capacity or the specific heat if the substance's mass is unity.

change in heat -

The amount of heat Q that is transported to create a temperature shift depends on the substance and phase involved, the mass of the system, and the magnitude of the temperature change. (A) The temperature change and the amount of heat transported are directly inversely related.

The answer's sign dig is determined by a calculation's lowest sign digits.

The solution will need 3 digits because the temperatures are provided with 3 significant digits and the specified heat has 3 significant digits (the mass has 4 significant digits).

31.91 g   * 0.903 J/g-C  *  (26.1 - 21.2 C) = 141 J

to learn more about Specific Heat go to - brainly.com/question/21406849

#SPJ4

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Answer:

<h3>The answer is 19,800 J</h3>

Explanation:

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

<h3>workdone = force × distance</h3>

From the question

force = 120 N

distance = 165 m

We have

work done = 120 × 165

We have the final answer as

<h3>19,800 J</h3>

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Social-cultural perspective

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8 0
3 years ago
You have landed on an unknown planet, Newtonia, and want to know what objects will weigh there. You find that when a certain too
steposvetlana [31]

Answer:

w'=5.679\ N on the planet

w=22.43\ N on earth

Explanation:

Given:

  • initial velocity of the tool before pushing, u=0\ m.s^{-1}
  • force applied on the tool, F=12\ N
  • displacement of the tool, s=16.4\ m
  • time taken for the displacement, t=2.5\ s
  • height of releasing  the tool, h=10.3\ m
  • time taken by the tool to fall on the ground, t_v=2.88\ s

<u>Now using the equation of motion:</u>

s=u.t+\frac{1}{2}a.t^2

where:

a = acceleration of the object

16.4=0+0.5\times a\times 2.5^2

a=5.248\ m.s^{-2}

Now the mass of the tool:

m=\frac{F}{a}

m=\frac{12}{5.248}

m=2.2866\ kg

<u>Using the equation of motion when the tool is dropped:</u>

h=u.t_v+\frac{1}{2} \times g.t_v^2

here:

g = acceleration due to gravity on the planet

10.3=0+0.5\times a\times 2.88^2

g=2.4836\ m.s^{-2}

Weight of the tool in the planet:

w'=m.g

w'=2.2866\times 2.4836

w'=5.679\ N

Weight of the tool on the earth:

w=m.g'

w=2.2866\times 9.81

w=22.43\ N

8 0
4 years ago
Read 2 more answers
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