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3 years ago

NEED HELP ASAP PLEASE

Physics

1 answer:

kompoz [17]3 years ago

8 0

did you end up getting the answer- im struggling

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The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth

ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

So, For finding gauge pressure we have formula P= ρ * g * h

Also gravity also remains same for both liquids

So taking ratio of their respective pressures we have

$\frac{P_{1} }{\\P_2}$ = $\frac{density * g * h_1}{density * g * h_2}$

So $\frac{39}{P_2}$ = $\frac{3}{9}$

Or P₂= 39 * 3 = 117 kPa

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3 years ago

Isabela creates a demonstration showing a sound echo. How could she demonstrate this?

jeka94

Answer:

Explanation:

took it in test prep

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2 years ago

A small boat coasts at constant speed under a bridge. A heavy sack of sand is dropped from the bridge onto the boat. The speed o

Tju [1.3M]

Answer:

d. decreases

Explanation:

The law of conservation of momentum tells us that the sum of momenta before the collision is equal to the sum of momenta after the collision. The bag has no momentum as it falls onto the boat because its velocity is zero in the horizontal direction. But after it hits the boat, it's momentum increases while the momentum of the system remains the same. That means a component of the system must decrease somewhere else. And that component is the velocity, not the mass, of the boat.

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4 years ago

A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0

barxatty [35]

Answer:

Velocity = 3.25[m/s]

Explanation:

This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.

In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.

The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"

h2 is located at point 2 and it will be zero.

$(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]$

4 0

4 years ago

Please i need help! ill give brainliest toooo!

Igoryamba

Answer:

Explanation:

7 0

3 years ago

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