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leonid [27]
3 years ago
11

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.1

10 m3 of air at a pressure of 0.355 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.390 m3 . If the temperature remains constant, what is the final value of the pressure?
Physics
1 answer:
Shkiper50 [21]3 years ago
5 0

Answer:

0.10013 atm

Explanation:

Applying Boyle's Law,

P'V' = PV................... Equation 1

Where P' = Initial pressure of air, V' = Initial volume of air, P = Final pressure of air, V = Final volume of air.

make P the subject of the equation

P = P'V'/V..................... Equation 2

Given: P' = 0.355 atm, V'  0.110 m³, V = 0.390 m³

Substitute into equation 2

P = 0.355(0.11)/0.39

P = 0.10013 atm.

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A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.
LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

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Distance from the center of platform = 1.07 m

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MOI of disk = \frac{MR^2}{2}

Moment of inertia of the person and the dog will be mr^2.

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Moment of inertia (I_y_y )  of the system with respect to the axis yy.

⇒ I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g

⇒ I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2

⇒ I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2

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The moment of inertia of the system is 289.088 kg.m^2

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