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3 years ago

4. In a object filled with a type of gas, which of the following actions would decrease the pressure

Physics

1 answer:

ExtremeBDS [4]3 years ago

7 0

Answer:

Decrease the number of gas particles, increase the object's area, and reduce the temperature of the gas

Explanation:

TEMPERATURE:

Decreasing the temperature will slow down the molecules. Hence, less no. of collisions will take place between walls of object and molecules. This will result in decrease of pressure.

Therefore, the pressure of a gas can be decreased by increasing its temperature.

NUMBER OF GAS PARTICLES:

Decreasing the number of particles will result in less no. of collisions, hence decreasing the pressure.

Therefore, the pressure of a gas can be decreased by decreasing its no. of molecules or no. of particles.

AREA OF OBJECT:

The pressure is given by the formula:

$P = \frac{F}{A}\\\\For constant Force (F):\\P\ \alpha\ \frac{1}{A}$

where,

A = Area of Object

Therefore, the pressure of a gas can be decreased by increasing area of object.

So, the correct option is:

Decrease the number of gas particles, increase the object's area, and reduce the temperature of the gas

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A block moves up a 30 degree incline under the action of three applied forces. F1 is horizontal and of magnitude 40 N. F2 is nor

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Answer

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A bowling pin is thrown vertically upward such that it rotates as it moves through the air, as shown in the figure. Initially, t

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Answer:

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3 years ago

A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one

alekssr [168]

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R= $13\Omega$

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper wire=Sum of resistance of two piece of wires

$r+7r=13$

$8r=13$

$r=\frac{13}{8}$ ohm

Resistance of long piece of wire= $7\times\frac{13}{8}=\frac{91}{8}\Omega$

Resistance of short piece of wire = $\frac{13}{8}\Omega$

Resistivity of wire and cross section area of wire remains same .

Let L be the total length of wire and L' be the length of short piece of wire.

We know that

$R=\frac{\rho L}{A}$ = $\frac{\rho}{A}L=KL$

$\frac{R}{L}=K$

Where K= $\frac{\rho}{A}$ =Constant

Using the formula

$\frac{13}{L}=\frac{\frac{13}{8}}{L'}$

$\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}$

$L'=\frac{L}{8}$

Length of short piece of wire=L'= $\frac{L}{8}$

Length of long piece of wire= $L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L$

% of length of short piece of wire= $\frac{\frac{L}{8}}{L}\times 100=12.5%$ %

The resistance of the short piece= $\frac{13}{8}\Omega$

The resistance of the long piece= $\frac{91}{8}\Omega$

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