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3 years ago

Compare these two waves.

Physics

1 answer:

Anastaziya [24]3 years ago

5 0

Answer:

id 272 966 3863

Pas 161005

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What are 3 types of contact forces

Svetach [21]

There are different types of contact forces like normal Force, spring force, applied force and tension force.

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3 years ago

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According to the law of conservation of mass, what is the mass of water

DENIUS [597]

Answer:

according to the conservation of mass,

according to the conservation of mass,the mass of the water is 36.04grams 

Explanation:

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3 years ago

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HURRY!

arsen [322]

Answer:

changes direction

changes speed

bounces off the boundary

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3 years ago

The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the ma

Law Incorporation [45]

Answer:

The speed of the particle is 2.86 m/s

Explanation:

Given;

radius of the circular path, r = 2.0 m

tangential acceleration, $a_t$ = 4.4 m/s²

total magnitude of the acceleration, a = 6.0 m/s²

Total acceleration is the vector sum of tangential acceleration and radial acceleration

$a = \sqrt{a_c^2 + a_t^2}\\\\$

where;

$a_c$ is the radial acceleration

$a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2$

The radial acceleration relates to speed of particle in the following equations;

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

where;

v is the speed of the particle

$v^2 = a_c r\\\\v= \sqrt{a_c r} \\\\v = \sqrt{4.08 *2}\\\\v = 2.86 \ m/s$

Therefore, the speed of the particle is 2.86 m/s

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4 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.

8 0

4 years ago