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The gravity of Earth is attracting a person towards the center with 500N of gravitational force. The person is exerting a reacti

mestny [16]

Answer:

500n

Explanation:

7 0

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

Art [367]

The total gauge pressure at the bottom of the cylinder would simply be the sum of the pressure exerted by water and pressure exerted by the oil.

The formula for calculating pressure in a column is:

P = ρ g h

Where,

P = gauge pressure

ρ = density of the liquid

g = gravitational acceleration

h = height of liquid

Adding the two pressures will give the total:

P total = (ρ g h)_water + (ρ g h)_oil

P total = (1000 kg / m^3) (9.8 m / s^2) (0.30 m) + (900 kg / m^3) (9.8 m / s^2) (0.4 - 0.30 m)

P total = 2940 Pa + 882 Pa

P total = 3,822 Pa

Answer:

The total gauge pressure at the bottom is 3,822 Pa.

6 0

3 years ago

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Pls help with either of these I will give up brainliest

Contact [7]

Answer:

1. The bird close to the center

2. 4/25 of the original force.

Explanation:

1. Tangential velocity is v=w*d (in m/s), where w is the rotational speed, commonly denoted as the letter omega (in radians per second). d is the distance from the center of the rotating object to the position of where you would like to calculate the velocity (in meters).

As we can note, the furthest from the center we are calculating the velovity the higher it is, because the rotational velocity is not changing but the distance of the object with respect to the center is. If v=w*d, then the lower the d (distance) the lower the tangential velocity.

2. Take a look at the picture:

We have the basic equation for the gravitational force.

We have to forces: Fg1, which is the original force, and Fg2, the force when the mass and the distance changes.

If we consider that mass 2 didn't change (m2'=m2), mass 1 is four times its original (m1'=4*m1) and distance is 5 times the original (r'=5*r), then next step is just plugging it into the equation for Fg2.

Dividing the original force Fg1 by the new force Fg2 (notice you can just as well do the inverse, Fg2 divided by Fg1) gives us the relation between the forces, cancelling all the variables and being left only with a simple fraction!