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lina2011 [118]
3 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]3 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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The pump in a water tower raises water with a density of rhow = 1.00 kg/liter from a filtered pool at the base of the tower up h
oee [108]

Answer:

Explanation:

mass of the water being lifted per second

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3 years ago
A 47.5-turn circular coil of radius 5.25 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0
Free_Kalibri [48]

Answer:

Torque, 5.91\times 10^{-3}\ N-m

Explanation:

Given that,

The number of turns in the coil, N = 47.5

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Current in the coil, I = 26.9 mA

The magnitude of the maximum possible torque exerted on the coil isg given by :

\tau=NIAB\\\\\tau=47.5\times 26.9\times 10^{-3}\times \pi (5.25\times 10^{-2})^2\times 0.535 \\\\\tau=0.00591\ N-m\\\\\tau=5.91\times 10^{-3}\ N-m

So, the magnitude of the maximum possible torque exerted on the coil is 5.91\times 10^{-3}\ N-m.

3 0
3 years ago
A trough in a transverse wave corresponds to a ________ <br> In a longitudinal wave.
kakasveta [241]

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5 0
2 years ago
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The heating element in a kettle behaves like a resistor. A particular kettle needs to operate at 230 V, with a power of 1500 W.
dedylja [7]

Answer:

R = 35.27 Ohms

Explanation:

Given the following data;

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To find the resistance, R;

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Where:

V is the voltage measured in volts.

R is the resistance measured in ohms.

Substituting into the equation, we have;

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Cross-multiplying, we have;

1500R = 52900

R = 52900/1500

R = 35.27 Ohms.

Therefore, the resistance which the heating element needs to have​ is 35.27 Ohms.

4 0
3 years ago
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