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3 years ago

If we warm a volume of air, it expands. Does it then follow that if we expand a volume of air, it warms? Explain.

Physics

2 answers:

wlad13 [49]3 years ago

7 0

Answer:

If the pressure is maintained at a constant value then both statements are equivalent:

P V = n R T Ideal gas equation

if P, n, and R are maintained at constant values then

V = T and the two expressions are equivalent

Ivahew [28]3 years ago

3 0

Answer:

nope don't think so

Explanation:

the heat causes the molecules to move faster therefore expanding in watever it the air is in

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A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

3 years ago

The electric field direction is defined by the direction of the force felt by (select one of the following answers):A. A negativ

steposvetlana [31]

Answer:

Explanation:

The formula for the electric field is Force (N)/charge(Coulombs). The electric field direction is defined by the direction of the force felt by a positive charge.

4 0

3 years ago

What is the middle layer of earth called

irinina [24]

The middle or centre of the Earth is the core. However the middle of the layers from the surface to the centre of the Earth is known as mantle.

6 0

3 years ago

Read 2 more answers

REAL ANSWER = BRAINLIST LOOK AT PICS ATTACHED

inysia [295]

Answer:

there are no pictures attached.

5 0

3 years ago

A student witnesses a flash of lightning and then t=2.5s later the student hears assiciated clap of thunder. (please show work)

frosja888 [35]

Answer:

857.5 m

2.8583×10⁻⁶ seconds

Explanation:

Time taken by the sound of the thunder to reach the student = 2.5 s

Speed of sound in air is 343 m/s

Speed of light is 3×10⁸ m/s

Distance travelled by the sound = Time taken by the sound × Speed of sound in air

⇒Distance travelled by the sound = 2.5×343 = 857.5 m

⇒Distance travelled by the sound = 857.5 m

Time taken by light = Distance the light travelled / Speed of light

$\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow \text{Time taken by light}=2.8583\times 10^{-6}$

Time taken by light = 2.8583×10⁻⁶ seconds

3 0

3 years ago