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3 years ago

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A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t

he total liquid depth is 40 cm. Part A What is the gauge pressure at the bottom of the cylinder

1 answer:

Anastaziya [24]3 years ago

6 0

Answer:

The gauge pressure is $P_g = 2058 \ P_a$

Explanation:

From the question we are told that

The height of the water contained is $h_w = 30 \ cm = 0.3 \ m$

The height of liquid in the cylinder is $h_t = 40 \ cm = 0.4 \ m$

At the bottom of the cylinder the gauge pressure is mathematically represented as

$P_g = P_w + P_o$

Where $P_w$ is the pressure of water which is mathematically represented as

$P_w = \rho_w * g * h_w$

Now $\rho_w$ is the density of water with a constant values of $\rho_w = 1000 \ kg /m^3$

substituting values

$P_w = 1000 * 9.8 * 0.3$

$P_w = 2940 \ Pa$

While $P_o$ is the pressure of oil which is mathematically represented as

$P_o = \rho_o * g * (h_t -h_w )$

Where $\rho _o$ is the density of oil with a constant value

$\rho _o = 900 \ kg / m^3$

substituting values

$P_o = 900 * 9.8 * (0.4 - 0.3)$

$P_o = 882 \ Pa$

Therefore

$P_g = 2940 - 882$

$P_g = 2058 \ P_a$

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3 years ago

In an elastic collision, a 300 kg bumper car collides directly from behind with a second, identical bumper car that is traveling

evablogger [386]

Answer:

If we had:

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Explanation:

In an elastic collision both linear momentum and kinetic energy are conserved, so we will have:

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We will call our bumpers 1 and 2.

For the momentum equation we know that:

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For the kinetic energy equation we know that:

$\frac{m_1v_{1i}^2}{2}+\frac{m_2v_{2i}^2}{2}=\frac{m_1v_{1f}^2}{2}+\frac{m_2v_{2f}^2}{2}$

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$v_{1i}+v_{2i}=v_{1f}+v_{2f}$

$v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2$

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$v_{1i}-v_{1f}=v_{2f}-v_{2i}$

$v_{1i}^2-v_{1f}^2=v_{2f}^2-v_{2i}^2$

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$(v_{1i}+v_{1f})(v_{1i}-v_{1f})=(v_{2f}+v_{2i})(v_{2f}-v_{2i})$

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$v_{1i}-v_{1f}+v_{2i}=v_{2f}$

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$v_{2i}+v_{2i}=+v_{1f}+v_{1f}$

$v_{1f}=v_{2i}$

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$v_{2f}=v_{1i}-v_{1f}+v_{2i}=v_{1i}-v_{1f}+v_{1f}=v_{1i}$

So we conclude that the bumpers have just exchanged velocities (something sometimes seen in billiards for example):

$v_{1f}=v_{2i}=5.9m/s$

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4 years ago