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3 years ago

The distance light travels in a year is called _______.

Physics

2 answers:

MrRissso [65]3 years ago

4 0

The answer would be a Light year.

lesya [120]3 years ago

4 0

the answer is a, a light year. Hope this helped!

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A stationary siren emits sound of frequency 1000 Hz and wavelength 0.343 m. An observer who is moving toward the siren will meas

Ainat [17]

Answer:

f>1000Hz and wavelength=0.343 m

Explanation:

We are given that

Frequency of stationary siren,f=1000 Hz

Wavelength of stationary sound, $\lambda=0.343 m$

When a observer is moving towards the siren then the frequency increases.

Therefore,an observer who is moving towards the siren measure a frequency >1000 Hz.

The wavelength depends upon the speed of source.

But we are given that siren is stationary.

Therefore, source is not moving and then the wavelength remains same.

f>1000Hz and wavelength=0.343 m

4 0

3 years ago

A person is riding on a Ferris wheel. When the wheel makes one complete turn, the net work done on the person by the gravitation

lara31 [8.8K]

Answer:

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

d = Vertical height from the ground

F = Force = Weight = mg

Net work done would be

$W_n=W_{up}+W_{down}\\\Rightarrow W_n=Fdcos180+Fdcos0\\\Rightarrow W_n=-mgd+mgd\\\Rightarrow W_n=0$

Hence, the work done on the person by the gravitational force is 0

7 0

3 years ago

When a loose brick is resting on a wall, it has energy. When the brick is pushed off the wall and is falling down, the amount of

OLga [1]

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3 years ago

Read 2 more answers

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

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3 years ago

While hovering motionless 3.0 meters an asteroid, a small

AnnZ [28]

I have no idea what you are trying to ask sorry

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3 years ago