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grigory [225]
3 years ago
9

A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume

that the pressure is constant and find its value.
Physics
1 answer:
Assoli18 [71]3 years ago
3 0

Answer:

Hence the pressure is 3\times 10^5 Pa

Explanation:

Given data

Q=1500 J   system gains heat

ΔV=- 0.010 m^3     there is a decrease in volume

ΔU= 4500 J        internal energy decrease

We know work done is

W= Q- ΔU

=1500-4500= -3000 J

The change in the volume at constant pressure is

ΔV= W/P

there fore P = W/ΔV= -3000/-0.01= 3×10^5

Hence the pressure is 3\times 10^5 Pa

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Answer:

1.6 J

Explanation:

Work = change in energy

W = ΔKE

Fd = KE

(0.8 N) (2 m) = KE

KE = 1.6 J

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Why is traveling between the stars (by creatures like us) difficult?
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Answer:

Stars are at large distances that even light takes thousands of years to reach us from there.

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4 0
3 years ago
A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te
Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

\Delta L=\alpha L_{0}\Delta T

Where:

  • L(0) is the initial length
  • ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
  • ΔL is the final length minus the initial length (L(f)-L(0))
  • α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)  

So, we have:

L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})

L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})

L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)

L_{f}=0.117\: m

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

7 0
3 years ago
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The relationship between frequency and wavelength for an electromagnetic wave is
c=f \lambda
where
f is the frequency
\lambda is the wavelength
c=3 \cdot 10^8 m/s is the speed of light.

For the light in our problem, the frequency is f=1.20 \cdot 10^{13} s^{-1}, so its wavelength is (re-arranging the previous formula)
\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{1.20 \cdot 10^{13} s^{-1}}=  2.5 \cdot 10^{-5}m
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