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BigorU [14]
3 years ago
10

what happens to the average molecular speed in an ideal gas if you triple the volume and keep the same temperature ?

Physics
1 answer:
kipiarov [429]3 years ago
4 0

Answer:

If you hold the temperature of an ideal gas constant, what happens to its volume when you triple the pressure? For T fixed, P is proportional to 1/V or V is proportional to 1/P. Tripling P reduces V to 1/3. ... If T is constant, the speeds of the average speeds and kinetic energy of the atomic particles remain constant.

I hope this helps!

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Which of these galaxies is most likely to be oldest?
ycow [4]

Answer:

a galaxy observed at a distance of 5 billion light-years

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2 years ago
Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q
katen-ka-za [31]

Answer:

c) 4.2*10^{-5}C

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

F=\frac{kq_{1}q_{2}}{r^{2}}

where k is a proportionality constant with the value k=9*10^{9}\frac{Nm^{2}}{C^{2}}

In this case q_{1}=q_{2}=q, so we have:

F=\frac{kq^{2}}{r^{2}}

Solving the equation for q, we have:

kq^{2}=Fr^{2}

q^{2}=\frac{Fr^{2}}{k}

q=\sqrt{\frac{Fr^{2}}{k}}

Replacing the given values:

q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}

q=4.2*10^{-5}C

3 0
3 years ago
In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera
Talja [164]

Answer:

η = 40 %  

Explanation:

Given that

Qa ,Heat addition= 1000  J

Qr,Heat rejection= 600 J

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We know that ,efficiency of a engine given as

\eta=\dfrac{W(net)}{Q(heat\ addition)}

Now by putting the values in the above equation ,then we get

\eta=\dfrac{400}{1000}

η = 0.4

The efficiency in percentage is given as

η = 0.4  x 100 %

η = 40 %

Therefore the answer will be 40%.

4 0
4 years ago
Is a process that modifies light waves so they vibrate in a single plane.
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4 0
4 years ago
Read 2 more answers
A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux
Mila [183]

Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

\Phi = EA cos \theta

where

E is the electric field

A is the cross-sectional area

\theta is the angle between the direction of the electric field and the normal to A

The flux is maximum when \theta=0^{\circ}, so we are in this situation and therefore cos \theta =1, so we can write

\Phi = EA

Here we have:

\Phi = 7.50\cdot 10^5 N/m^2 C is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2

And so, we can find the magnitude of the electric field:

E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C

3 0
4 years ago
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