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3 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]3 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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Explanation:

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3 years ago

Compare these two collisions of a PE student with a wall.

Stolb23 [73]

1) The variable that is different in the two cases is $\Delta t$ , the duration of the collision

2) The change in momentum is the same in the two cases

3) The impulse is the same in the two cases

4) Case B will experience a greater force

Explanation:

The variable that is different in the two cases is $\Delta t$ , the duration of the collision.

In fact, in the first case the wall is padded: this means that the collision will be "softer" and therefore will last longer, so the duration of the collision, $\Delta t$ , will be larger.

In the second case instead, the wall is unpadded: this means that the collision is "harder" and so it will last less time, therefore the duration of the collision $\Delta t$ will be smaller.

The change in momentum in the two cases is the same.

In fact, the change in momentum is given by:

$\Delta p = m(v-u)$

where:

m is the mass of the student

u is the initial velocity

v is the final velocity

In both cases, we have:

m = 75 kg

u = 8 m/s

v = 0 (they both comes to rest)

Therefore, the change in momentum is

$\Delta p = (75)(0-8)=-600 kg m/s$

The impulse in the two cases is the same.

In fact, impulse is defined as the product of force applied, F, and duration of the collision, $\Delta t$ :

$J=F \Delta t$

However, the force can be rewritten as product of mass (m) and acceleration (a), according to Newton's second law:

$F=ma$

So the impulse is

$J=ma\Delta t$

The acceleration can be rewritten as rate of change of velocity:

$a=\frac{\Delta v}{\Delta t}$

So the impulse becomes

$J=m\frac{\Delta v}{\Delta t}\Delta t = m\Delta v$

So, the impulse is equal to the change in momentum: and since in the two cases the change in momentum is the same, the impulse is the same as well.

The force in the collision is related to the impulse by

$J=F\Delta t$

where

J is the impulse

F is the force

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The equation can be rewritten as

$F=\frac{J}{\Delta t}$

In the two situations described in the problem (A and B), we already said that the impulse is the same (because the change in momentum is the same). However, in case A (padded wall) the time $\Delta t$ is longer, while in case B (unpadded wall) the time $\Delta t$ is shorter: since the force F is inversely proportional to the duration of the collision, this means that in case B the student will experience a greatest force compared to case A.

Learn more about impulse:

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3 years ago

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zhenek [66]

Answer:

F = 50[N], to the left.

v = 10.52 [m/s]

Explanation:

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In order to solve this problem we must apply Newton's laws, in such a way that we must perform a summation of forces on the horizontal axis. In this way we will analyze each force and the direction of action.

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100 - 150 = F

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Second problem

We must remember that the definition of speed is equal to the relationship between distance over time.

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t = time = 9.5 [s]

v = x/t

v = 100/9.5

v = 10.52 [m/s]

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4 years ago

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