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anzhelika [568]
3 years ago
12

A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

eased to 125 kPa if the temperature remains constant? *
Physics
1 answer:
const2013 [10]3 years ago
7 0

Answer:

388 cm^3

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

pV=const.

which can also be rewritten as

p_1 V_1 = p_2 V_2

In our case, we have:

p_1 = 75.9 kPa is the initial pressure

V_1 = 639 cm^3 is the initial volume

p_2 = 125 kPa is the final pressure

Solving for V2, we find the final volume:

v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3

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We have an object with a density of 620g/cm^3 and a volume of 75cm cm^3. What is the mass of this object
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3 years ago
A baseball has a mass of 0.15 kg and radius 3.7 cm. In a baseball game, a pitcher throws the ball with a substantial spin so tha
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Rotational kinetic energy = 0.099 J

Translational kinetic energy = 200 J

The moment of inertia of a solid sphere is I = \frac{2}{5}mr^2.

Explanation:

Rotational kinetic energy is given by

\text{RKE} = \frac{1}{2}I\omega^2

where <em>I</em> is the moment of inertia and <em>ω</em> is the angular speed.

For a solid sphere,

I = \frac{2}{5}mr^2

where <em>m</em> is its mass and <em>r</em> is its radius.

From the question,

<em>ω</em> = 49 rad/s

<em>m</em> = 0.15 kg

<em>r</em> = 3.7 cm = 0.037 m

\text{RKE} = \frac{1}{2}\times \frac{2}{5} mr^2\omega^2 = \frac{1}{5} mr^2\omega^2

\text{RKE} = \frac{1}{5} (0.15\ \text{kg})(0.037\ \text{m})^2(49\ \text{rad/s})^2 = 0.099\text{ J}

Translational kinetic energy is given by

\text{TKE} = \frac{1}{2} mv^2

where <em>v</em> is the linear speed.

\text{TKE} = \frac{1}{2} (0.15\ \text{kg})(52\ \text{m/s})^2 = 200\text{ J}

5 0
3 years ago
Read 2 more answers
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