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3 years ago

Rays traveling parallel to the principle axis of a concave mirror will reflect out through the mirrors focus?

Physics

2 answers:

abruzzese [7]3 years ago

7 0

Answer:

True

Explanation:

When a ray travelling parallel to the principle axis of a concave mirror then the light ray reflect out through the mirrors and passing through the focus.

When a light ray travelling through focus of a concave mirror then after reflection the light ray reflect out through the mirror and go parallel to principle axis.

Therefore, rays travelling parallel to the principle axis of a concave mirror will reflect out through the mirrors focus.

It is true.

kompoz [17]3 years ago

7 0

Answer:

true

Explanation:

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The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surfac

andrezito [222]

Answer:

$E=8.13\times 10^{12}\ J$

Explanation:

Given that,

The mass of a Hubble Space Telescope, $m_1=1.16\times 10^4\ kg$

It orbits the Earth at an altitude of $5.68\times 10^5\ m$

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

$E=\dfrac{Gm_1m_e}{r}$

Where

$m_e$ is the mass of Earth

Put all the values,

$E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J$

So, the potential energy of the telescope is $8.13\times 10^{12}\ J$ .

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3 years ago

Can someone answer all of the questions?

Katyanochek1 [597]

Answer: -200

Explanation: is it math

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3 years ago

Read 2 more answers

What is the radius of the Earth?

ki77a [65]

<span>the answer would be 3,959 miles</span>

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3 years ago

A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun

cluponka [151]

Answer:Highest frequency =618.89Hz

Lowest frequency=582.22Hz

Explanation:

The linear velocity of a sound generator is related to angular velocity and is given as

Vs = rω where

r = the radius of circular path = 1.0 m

ω is the angular velocity of the sound generator. = 100 rpm

1 rev/min = 0.10472 rad/s

100rpm =10.472 rad/ s

Vs = rω

= 1m x 10.472rad/ s= 10.472m/s

A) Highest frequency heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,

f max = (v/ v-vs) fs

Where , v is the speed of the sound in air at 20 degrees celcius =

343 metres per second

vs is the linear velocity of the sound generator=10.472m/s

fs is the frequency of the sound generator= 600 Hz

f max = (343/ 343 - 10.472) x 600

=343/332.528) x600

=618.89Hz

B) Lowest frequency heard by a student in the classroom = Minimum frequency

f min = (v/ v+vs) fs

(343/ 343 + 10.472) x 600

=343/353.472) x 600

=582.22hz

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3 years ago

Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea

Kipish [7]

Answer:

6.5 m/s

Explanation:

We are given that

Distance, s=100 m

Initial speed, u=1.4 m/s

Acceleration, $a=0.20 m/s^2$

We have to find the final velocity at the end of the 100.0 m.

We know that

$v^2-u^2=2as$

Using the formula

$v^2-(1.4)^2=2\times 0.20\times 100$

$v^2-1.96=40$

$v^2=40+1.96$

$v^2=41.96$

$v=\sqrt{41.96}$

$v=6.5 m/s$

Hence, her final velocity at the end of the 100.0 m=6.5 m/s

5 0

3 years ago