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2 years ago

What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.

Physics

1 answer:

12345 [234]2 years ago

5 0

Incomplete question.The complete question is

A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm. What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.

Answer:

Total gauge pressure= 3724Pa or 3.724kPa

Explanation:

Pressure (water alone) = (1,000 x 9.8 x 0.2m) = 1,960Pa.

Pressure (oil on top) = (900 x9.8 x 0.2m) = 1,764Pa.

Total gauge pressure at bottom = (1,960 + 1,764) = 3724Pa or 3.724kPa

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The boiling point for liquid nitrogen at atmospheric pressure is 77k. Is the temperature of an open container of liquid nitrogen

algol13

Answer:
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3 years ago

a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik

grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height of the ball when it hits the ground

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball

$t$ is the time when the ball strikes the ground

$g=9.8m/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $t$ from (1):

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$ (5)

Solving (5) we find the positive result is:

$t=17.68 s$

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2 years ago

A ball rolls across a floor with an acceleration of 0.100 m/s2 in a direction opposite to its velocity. The ball has a velocity

Westkost [7]

Answer:

4.15 m/s

Explanation:

Its given that acceleration is 0.1 m/s² with a direction opposite to the velocity. Since, the direction of acceleration is opposite to the velocity, this gives us a hint that the velocity is decreasing and so acceleration would be negative.

i.e.

acceleration = a = - 0.1 m/s²

Distance covered = S = 6m

Velocity after covering 6 meters = Final velocity = $v_{f}$ = 4 m/s