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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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Mnenie [13.5K]
Agree

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4 0
3 years ago
Read 2 more answers
If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radia
Tanya [424]

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

3 0
3 years ago
Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
dezoksy [38]
The De Broglie wavelength of the electron is
\lambda=34.0 nm=34 \cdot 10^{-9} m
And we can use De Broglie's relationship to find its momentum:
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s

Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}=  2.13 \cdot 10^4 m/s

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J
3 0
3 years ago
Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin
likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0
3 years ago
A wave has a wavelength of 3.0 m, a frequency of 25.0 Hz, and an amplitude of 14.0 cm. The wave travels in the positive x-direct
Mrac [35]

Answer:

The total number of oscillations made by the wave during the time of travel is 1.4 Oscillations. Strictly speaking, the number of complete oscillations is 1.

Explanation:

The required quantity is the number of complete oscillations made by the traveling wave. The amplitude time and frequency are not needed to calculate the number of oscillations as it is the ratio of the distance traveled to the wavelength( minimum distance that must be traveled to complete one oscillation) of the wave. So the total number of oscillations is 1.4 while the number of complete oscillations is 1 (strictly speaking). The detailed solution to this question can be found in the attachment below. Thank you!

4 0
3 years ago
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