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lina2011 [118]
2 years ago
14

If we double the diameter of a telescope's mirror, what happens to its light-gathering ability? View Available Hint(s) If we dou

ble the diameter of a telescope's mirror, what happens to its light-gathering ability?
Light-gathering ability doubles as well.
Light-gathering ability quadruples.
There is no change.
The light-gathering ability is cut in half.
Physics
1 answer:
crimeas [40]2 years ago
7 0

Answer:

Light-gathering ability quadruples.

Explanation:

The light-gathering ability of a telescope depends on the area of the mirror, which is the surface presented to the light. Since these mirrors are circular (diameter is mentioned), we need to know what happens with the area of it when its diameter is doubled.

If r is the radius and d the diameter of a circle, its area will be:

A=\pi r^2=\pi (\frac{d}{2} )^2=\pi(\frac{d^2}{4})=\frac{\pi d^2}{4}

We want to know now what happens if we double the diameter to 2d, so we calculate the new area A':

A'=\frac {\pi (2d)^2}{4}=\frac {\pi 4d^2}{4}=\pi d^2

Which is 4 times greater than the original area A, so we conclude that its light-gathering ability quadruples.

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3 years ago
3 bulbs are in series and the same 3 bulbs are in parallel with the same battery. Which bulbs will be dimmer?
Marina86 [1]

3 bulbs are in series and if the same 3 bulbs are in parallel with the same battery then the bulbs that are connected in parallel  will be dimmer

<h3>What is power?</h3>

The rate of doing work is known as power. The Si unit of power is the watt.

Power =work/time

The mathematical expression for the electric power is as follows

P = VI  

The same current flows through both bulbs when they are connected in series. A greater voltage drop across the bulb with the higher resistance will result in higher power dissipation and brightness. In the case of the parallel combination, the bulb will be dimmer

Thus, If the same three bulbs are connected in series and parallel with the same battery, the parallelly connected bulbs will be dimmer, therefore the correct option is A

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3 0
1 year ago
An engine has an energy input of 125j and 35 j of that energy is transformed into useful energy
Tatiana [17]
The engine's efficiency is (35J)/125J) = 28% .

Do you have a different question to ask ?
5 0
2 years ago
what is the name of the tool that allows you to copy two or more shapes into a single part? when working with 3D
Anna [14]

Answer:

Modeling tool or Align tool. it depends what type of sandbox platform you use

Explanation:

1

8 0
2 years ago
An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place
anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

mu=(M+m)v

v=\frac{m}{M+m}\times u

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

\frac{(M+m)v^2}{2}=(M+m)gh

here h=L(1-\cos \theta ) from diagram

therefore

v=\sqrt{2gL(1-\cos \theta )}

initial velocity in terms of v

u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}

For first case \theta =6.8^{\circ}

u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}

for second case \theta =11.4^{\circ}

u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}

Therefore \frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}

\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}

\frac{u_1}{u_2}=1.084

i.e.\frac{v_1}{v_2}=1.084

4 0
3 years ago
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