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3 years ago

Factors that affect pressure in liquids

Physics

2 answers:

Alona [7]3 years ago

7 0

Two factors influence the pressure of fluids. They are the depth of the fluid and its density.

Maksim231197 [3]3 years ago

6 0

Answer:

Depth and density of the liquid determine the amount of pressure in a liquid.

Explanation:

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A 950kg elevator is lifted 15m at 1.0m/s2. calculate the work is done on the elevator

blagie [28]

There it is fella tried on ma own consciousness

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3 years ago

use the general formulas for gravitational force and centripetal force to derive the relationship between speed and orbital radi

MrRissso [65]

Answer:

$v = \sqrt{\frac{GM}{r}}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{Mm}{r^{2}}$ (1)

Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.

The centripetal force can be found by means of Newton's second law:

$F = ma$ (2)

Since it is a circular motion, the acceleration can be defined as:

$a = \frac{v^{2}}{r}$ (3)

Where v is the velocity and r is the orbital radius.

Replacing equation (3) in equation (2) it is gotten:

$F = m\frac{v^{2}}{r}$ (4)

Hence,

$m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}$

Then, v can be isolated:

$mv^{2} = G\frac{Mmr}{r^{2}}$

$mv^{2} = G\frac{Mm}{r}$

$v^{2} = G\frac{Mm}{mr}$

$v^{2} = \frac{GM}{r}$

$v = \sqrt{\frac{GM}{r}}$

So the relationship between speed and orbital radius is given by the expression $v = \sqrt{\frac{GM}{r}}$

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3 years ago

An astronomer observes a hydrogen line in the spectrum of a star. The wavelength of hydrogen in the laboratory is 6.563 x 10-7m,

Sergio [31]

Answer:

D. The star is approaching Earth.

Explanation:

As we know by the doppler's effect of light that

$\frac{\Delta \nu}{\nu} = \frac{v}{c}$

here we know that

$\Delta \nu$ = change in frequency

here we know that the wavelength of light coming from the star is decreased so the frequency will increase

$\Delta \nu = \frac{c}{\lambda'} - \frac{c}{\lambda}$

$\Delta \nu =(3\times 10^8)(\frac{1}{6.56186 \times 10^{-7}} - \frac{1}{6.563 \times 10^{-7}})$

$\Delta \nu = 7.9414 \times 10^{10} Hz$

now we have

$\frac{7.9414 \times 10^{10}}{\nu} = \frac{v}{c}$

here we know that

$\nu = \frac{3\times 10^8}{6.563 \times 10^{-7}} = 4.57 \times 10^{14} Hz$

now we have

$v = 5.2 \times 10^4 m/s$

So here correct answer is

D. The star is approaching Earth.

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Answer:

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