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diamong [38]
2 years ago
12

A gas occupies 2 m^3 at 27°C at a pressure of 1 atmosphere. At a pressure of 2 atmospheres it occupies a volume of 1 m^2. What i

s its temperature at this new volume and pressure?
Physics
1 answer:
diamong [38]2 years ago
6 0

Answer:

27°C

Explanation:

We'll begin by converting 27 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 27 °C

Initial temperature (T₁) = 27 °C + 273

Initial temperature (T₁) = 300 K

Next, we shall determine the final temperature of the gas. This can be obtained as follow:

Initial volume (V₁) = 2 m³

Initial temperature (T₁) = 300 K

Initial pressure (P₁) = 1 atm

Final pressure (P₂) = 2 atm

Final volume (V₂) = 1 m³

Final temperature (T₂) =?

P₁V₁/T₁ = P₂V₂/T₂

1 × 2 / 300 = 2 × 1 / T₂

2/300 = 2/T₂

1/150 = 2/T₂

Cross multiply

T₂ = 150 × 2

T₂ = 300 K

Finally, we shall convert 300 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 300 K

T(°C) = 300 – 273

T(°C) = 27°C

Thus, the final temperature is 27°C

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6 0
2 years ago
(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro
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Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:  

v = \frac{m + M}{m} \cdot \sqrt{2gh}

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.   </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:    

\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}           (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.  </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}  

\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}

\frac{v_{2}}{v_{1}} = 1.41  

Therefore, the second projectile was 1.41 times faster than the first.  

I hope it helps you!

8 0
3 years ago
PLEASE HELP QUICKLY THANKS
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Answer: the answer is d

Explanation:

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Answer:

\frac{F_1}{F_2} =\frac{1}{2}

Explanation:

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Case 1:

F_1=2k\times d

case 2:

F_2=2k\times 2d

therefore, \frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}

\frac{F_1}{F_2} =\frac{1}{2}

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