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Natalka [10]
3 years ago
7

The CEO, ellen misk, left her martian office but accidentally left a cylindricall can of coke (3.1 inches in diameter, 5.42 inch

es in height) on her desk. If the can exerts a pressure of 510 Pascals, what is the specific gravity of the can?
Physics
1 answer:
iren2701 [21]3 years ago
8 0

Answer:

Specific Gravity = 0.378

Explanation:

First, we will find the force exerted by the can on the table. This force will be equal to the weight of the can:

Pressure = Force/Area = Weight/Area

Weight = Pressure*Area

where,

Area = πdiameter²/4 = π[(3.1 in)(0.0254 m/1 in)]²/4 = 4.8 x 10⁻³ m²

Weight = (510 N/m²)(4.8 x 10⁻³ m²)

Weight = 2.48 N

Now, the weight is given as:

Weight = mg

2.48 N = m(9.8 m/s²)

m = (2.48 N)/(9.8 m/s²)

m = 0.25 kg

Now, we calculate volume of can:

Volume = (Area)(Height) = (4.8 x 10⁻³ m²)(5.42 in)(0.0254 m/1 in)

Volume = 6.6 x 10⁻⁴ m³

Hence, the density of can will be:

Density of Can = m/Volume = 0.25 kg/6.6 x 10⁻⁴ m³

Density of Can = 378.32 kg/m³

So, the specific gravity of Can will be:

Specific Gravity = Density of Can/Density of Water

Specific Gravity = (378.32 kg/m³)/(1000 kg/m³)

<u>Specific Gravity = 0.378</u>

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Answer:

\frac{v_{2}}{v_{1}}=2.

Explanation:

The average kinetic energy per molecule of a ideal gas is given by:

\bar{K}=\frac{3k_{B}T}{2}

Now, we know that \bar{K} = (1/2)m\bar{v}^{2}

Before the absorption we have:

(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2} (1)

After the absorption,

(1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2} (2)

If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}

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

\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}

\frac{v_{2}}{v_{1}}=\sqrt{4}

Therefore the ratio will be \frac{v_{2}}{v_{1}}=2

I hope it helps you!

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3 years ago
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earnstyle [38]

Answer:

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Angular acceleration is the time rate of change of angular velocity . In SI units, it is measured in radians per second squared (rad/s²)

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a = (2 - 4)/5 = -2/5 =

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\boxed{\large{\bold{\textbf{\textsf{{\color{blue}{Answer}}}}}}:)}

\sf{\qquad{\qquad{\underline{\underline{ Projectile~motion }}}}}

If an object is given an initial velocity in any direction and then allowed to travel freely under gravity only, it is called a projectile motion.

It is basically 3 types

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The path followed by a projectile is called its trajectory.

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Answer:

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