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den301095 [7]
3 years ago
5

In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2820 J of work is done on the gas

.
Required:
a. How much heat flows into or out of the gas?
b. What is the change in internal energy of the gas?
c. Does its temperature rise or fall?
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

Answer:

  1. Q=0
  2. U=2820
  3. Energy increases

Explanation:

From the question we are told that

Work done W=2820

a)Generally the heat flow for an adiabatic process is 0 (zero)

Q= U + W =>0

Q=0

b)Generally Change in internal energy of gas is mathematically given by

Since W=-2820J

Therefore

U=2820

Giving

Q= 2820 -2820

Q=O

c)With increases in internal energy brings increase in temperature

Therefore

Energy increases

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Answer:

The force required to push to stop the car is 288.67 N

Explanation:

Given that

Mass of the car, m = 1000 kg

Initial speed of the car, u = 1 m/s

The car and push on the hood at an angle of 30° below horizontal, \theta=30^{\circ}

Distance, d = 2 m

Let F is the force must you push to stop the car.

According work energy theorem theorem, the work done is equal to the change in kinetic energy as :

W=\dfrac{1}{2}m(v^2-u^2)F\times d=\dfrac{1}{2}m(v^2-u^2)

v = 0

Fd\ cos\theta=\dfrac{1}{2}m(u^2)      F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}F = -288.67 N

The force required to push to stop the car is 288.67 N

3 0
3 years ago
A ____mixture is made of different materials that can easily be separted
Firdavs [7]

Answer:

Heterogeneous

Explanation:

5 0
3 years ago
compare to visible light, the wavelength of x-rays is shorter longer or the same and the frequency is lower higher or the same
Doss [256]

Answer:shorter higher

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7 0
3 years ago
A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect
zaharov [31]

Answer:

X=X_o+\dfrac{1}{2}gt^2

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

\dfrac{1}{2}mV^2=mg(X-X_o)

So

V=\sqrt{2g(X-X_o)}

We know that

\dfrac{dX}{dt}=V

\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}

\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX

At t= 0 ,X=Xo

So we can say that

X=X_o+\dfrac{1}{2}gt^2

So the length of cable after time t

X=X_o+\dfrac{1}{2}gt^2

6 0
3 years ago
A spherical balloon is made from a material whose mass is 3.30 kg. The thickness of the material is negligible compared to the 1
stellarik [79]

Answer:

563712.04903 Pa

Explanation:

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r = Radius of sphere = 1.25 m

v = Volume of balloon = \frac{4}{3}\pi r^3

M = Molar mass of helium = 4.0026\times 10^{-3}\ kg/mol

\rho = Density of surrounding air = 1.19\ kg/m^3

R = Gas constant = 8.314 J/mol K

T = Temperature = 345 K

Weight of balloon + Weight of helium = Weight of air displaced

mg+m_{He}g=\rho vg\\\Rightarrow m_{He}=\rho vg-m\\\Rightarrow m_{He}=1.19\times \frac{4}{3}\pi 1.25^3-3.3\\\Rightarrow m_{He}=6.4356\ kg

Mass of helium is 6.4356 kg

Moles of helium

n=\frac{m}{M}\\\Rightarrow n=\frac{6.4356}{4.0026\times 10^{-3}}\\\Rightarrow n=1607.85489

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P=\frac{nRT}{v}\\\Rightarrow P=\frac{1607.85489\times 8.314\times 345}{\frac{4}{3}\pi 1.25^3}\\\Rightarrow P=563712.04903\ Pa

The absolute pressure of the Helium gas is 563712.04903 Pa

3 0
3 years ago
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