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MAVERICK [17]
3 years ago
12

The diffraction limit is a limit on: The diffraction limit is a limit on: A telescope's size. A telescope's angular resolution.

A telescope's spectral resolution.
Physics
1 answer:
Mnenie [13.5K]3 years ago
7 0

Answer:

A telescope's angular resolution.

Explanation:

Diffraction limit is a minimum angular separation of two sources and it can be distinguished by the telescope. This angle is known as the diffraction limit. It is proportional to the wavelength of light and it has an inverse relation with the diameter of the telescope. Mathematically  it is defined as

θ = 1.22λ/d

where θ is the angle, λ wavelength and d is the diameter of the objective mirror (lenz).

You might be interested in
An object takes 5 seconds to move 2 meters upward. How fast did it go?
jonny [76]

Answer:

2.5

Explanation:

5/2=2.5

8 0
3 years ago
A crane used 250,000 Joules of work to move a beam to the top of a building in 20 seconds. How much power did the crane use?
maria [59]

Answer:

12500W

Explanation:

Given parameters:

Work done  = 250000J

Time taken  = 20s

Unknown:

Power of the crane = ?

Solution:

Power is the defined as the rate at which work is being done;

  Mathematically;

        Power = \frac{work done}{time }

 insert the parameters and solve;

      Power  = \frac{250000}{20}   = 12500W

7 0
3 years ago
Two steamrollers begin 105 mm apart and head toward each other, each at a constant speed of 1.20 m/s. At the same instant, a fly
levacccp [35]

Answer: 109.4 mm

Explanation: <u>Distance</u> is a scalar quantity and it is the measure of how much path there are between two locations. It can be calculated as the product of velocity and time:  d = vt

The separation between the two steamrollers is 105 mm or 0.105 m. They collide to each other at the middle of the separation:

location of collision = \frac{0.105}{2} = 0.0525 m

To reach that point, both steamrollers will have spent

v=\frac{\Delta x}{t}

t=\frac{\Delta x}{v}

t=\frac{0.0525}{1.2}

t = 0.04375 s

The fly is travelling with speed of 2.5 m/s. So, at t = 0.04375 s:

d = 2.5*0.04375

d = 0.109375 m

Until it is crushed, the fly will have traveled 109.4 mm.

4 0
3 years ago
Robert bought 20 games for his grandkids. If he spent a total of $199.80, how much did each game cost if
RideAnS [48]

Answer:

$199.80/20 =,   the solution is 9.99 per game. Hope it helped! :D

Explanation:

4 0
3 years ago
10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr
12345 [234]

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

F_{g0}=G \frac{mM_{E0}}{r^{2}}

So if the mass of the earth is increased by a factor of 2, this means that:

M_{Ef}=2M_{E0}

so:

F_{gf}=G \frac{2mM_{E0}}{r^{2}}

Therefore:

\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}

When simplifying we end up with:

\frac{F_{gf}}{F_{g0}}=2

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

M_{Ef}=3M_{E0}

so:

F_{gf}=G \frac{3mM_{E0}}{r^{2}}

Therefore:

\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}

When simplifying we end up with:

\frac{F_{gf}}{F_{g0}}=3

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

M_{Ef}=\frac{M_{E0}}{4}

so:

F_{gf}=G \frac{mM_{E0}}{4r^{2}}

Therefore:

\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}

When simplifying we end up with:

\frac{F_{gf}}{F_{g0}}=\frac{1}{4}

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

4 0
3 years ago
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