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Kruka [31]
2 years ago
5

A piston absorbs 25 J of heat from its surroundings while expanding from

Physics
1 answer:
Ivahew [28]2 years ago
6 0
  • The work done is -30 J
  • The heat is 25 J

<h3>What is the heat and the work?</h3>

We know that the work done by a gas could be positive or negative same as the heat. If the work done is positive then work is done on the system.

The work done is obtained from;

W = PΔV

W =  1.0 x 105 Pa(0.0006 m³ - 0.0003 m³)

W = 30 J

Given that the gas absorbs heat from the surroundings and the gas is expanding.

  • The work done is -30 J
  • The heat is 25 J

Learn more about work done on a gas:brainly.com/question/12539457

#SPJ1

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A 1,400 kg car accelerates from rest to 30 m/s in 6.0 seconds. what is the net force on the car?
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So your finding acceleration first which is 30m/s divides by 6 seconds equals 5m/s^s and then multiply that by 1,400 kg and you have net force which is 7,000N
4 0
3 years ago
Determine the inductive reactance for a 50 mH inductor that is across a 15 volt, 400 Hz source.
tresset_1 [31]

Answer:

Inductive reactance is 125.7 Ω

Explanation:

It is given that,

Inductance, L=50\ mH=0.05\ H

Voltage source, V = 15 volt

Frequency, f = 400 Hz

The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :

X_L=2\pi fL

X_L=2\pi \times 400\ Hz\times 0.05\ H

X_L=125.66\ \Omega

X_L=125.7\ \Omega

So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.

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3 years ago
Describe a procedure that would increase the potential energy of two magnets if like poles are used. Explain why the energy of t
zalisa [80]

Answer:

If you apply a force to separate 2 opposite poles, the potential energy of the system increases.

5 0
2 years ago
Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw
Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of  the distance between them, acting along  the  line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the  factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0
3 years ago
In order to design an experiment you need a ____ about the scientific question you are trying to answer
s2008m [1.1K]

C Hypothesis. It is the correct answer.

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