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yuradex [85]
3 years ago
10

Suppose you have a piston in a cylinder where the gas temperature is 26.0° C and has a volume of 1.75 L. What is the volume of t

he gas if you move the assembly to a cooler region with a temperature of 3.00° C? Assume the gas pressure within the cylinder is constant. Express your answer in L.
Physics
2 answers:
blsea [12.9K]3 years ago
8 0

Answer:

1.62 L

Explanation:

Charle's law for ideal gases states that for a gas kept at constant pressure, the ratio between volume and temperature is constant:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where in this problem we have

V_1 = 1.75 L is the initial volume of the gas

T_1 = 26.0^{\circ}+273=299 K is the initial temperature of the gas

V_2 is the final volume of the gas

T_2 = 3.0^{\circ}+273=276 K is the final temperature

Solving the equation for V2, we find

V_2 = \frac{V_1}{T_1} T_2 = \frac{1.75 L}{299 K}(276 K)=1.62 L

TEA [102]3 years ago
4 0

what was the answer

Explanation:

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A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

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