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2 years ago

How much work is done by 0.070 m3 of gas, when the volume remains constant with pressure of 63 x 105 Pa?

Physics

1 answer:

stealth61 [152]2 years ago

5 0

Answer:

W = 0 J

Explanation:

The amount of work done by gas at constant pressure is given by the following formula:

$W = P\Delta V$

where,

W = Work done by the gas

P = Pressure of the gas

ΔV = Change in the volume of the gas

Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:

$W = P(0\ m^3)\\$

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A transverse wave on a string has an amplitude a. A tiny spot on the string is colored red. As one cycle of the wave passes by

aliya0001 [1]

Answer:

Option D) 4A

Explanation:

As the cycle of the wave passes by, the amplitude gives the longest journey when the spot travels from the undistributed position. During each cycle the spot travels "Four times" .

Considering one of this cycle, if it begins to travel from it's undistributed position , there would be four movements i.e

* Upward movement through distance A

*Downward movement through distance A

*Downward again through distance A

*Upward through distance A.

Then it would travel back to its undistributed position held

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2 years ago

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Alexeev081 [22]

B student 2 because you add

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2 years ago

A wooden block has a mass of 562 g and a volume of 72 cm3. what is the density

Gnom [1K]

Answer:

7 $\frac{29}{36}$ g/cm³

Explanation:

The formula for density is $\frac{mass}{volume}$ . Let's apply the formula to the question:

$\frac{562}{72}$ = 7 $\frac{29}{36}$ g/cm³

7 0

2 years ago

In an experiment conducted using the Young's double-slit apparatus, the separation between the slits is 20 µm. A first-order con

jeka57 [31]

To solve this problem, we use the formula
λ = s sin θ
where s is the separation and θ is the angle interference
So,
λ = 20 x 10^-6 sin 2.5
λ = 8.72 x 10^-7 m

The required angle for the fourth order bright fringe is
θb = sin⁻¹ (4λ / s) = sin⁻¹ (4 (8.72 x 10^-7 m)/ 20 x 10^-6 ) = 10.04°

The required angle for the fourth order dark fringe is
θd = sin⁻¹ (4.5 λ / s) = sin⁻¹ (4.5 (8.72 x 10^-7 m)/ 20 x 10^-6 ) = 11.31°

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2 years ago

A sample of gas with a volume of 750 ml exerts a pressure of 98 kpa at 30◦c. What pressure will the sample exert when it is comp

Tanzania [10]

Answer:

241 kPa

Explanation:

The ideal gas law states that:

$pV=nRT$

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

We can rewrite the equation as

$\frac{pV}{T}=nR$

For a fixed amount of gas, n is constant, so we can write

$\frac{pV}{T}=const.$

Therefore, for a gas which undergoes a transformation we have

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where the labels 1 and 2 refer to the initial and final conditions of the gas.

For the sample of gas in this problem we have

$p_1 = 98 kPa=9.8\cdot 10^4 Pa\\V_1 = 750 mL=0.75 L=7.5\cdot 10^{-4}m^3\\T_1 = 30^{\circ}C+273=303 K\\p_2 =?\\V_2 = 250 mL=0.25 L=2.5\cdot 10^{-4} m^3\\T_2 = -25^{\circ}C+273=248 K$

So we can solve the formula for $p_2$ , the final pressure:

$p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(7.5\cdot 10^{-4} m^3)(248 K)}{(303 K)(2.5\cdot 10^{-4} m^3)}=2.41\cdot 10^5 Pa = 241 kPa$

4 0

2 years ago