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gizmo_the_mogwai [7]
3 years ago
5

If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resol

ved if their angular separation is 3.0×10−5 radians? Express your answer in nano.
Physics
1 answer:
Alex3 years ago
6 0

This question is incomplete.Here is complete one

Consider a telescope with a small circular aperture of diameter 2.0 centimeters. If two point sources of light are being imaged by this telescope, what is the maximum wavelength (lambda) at which the two can be resolved if their angular separation is 3.0x10^-5 radians? Express your answer in nanometers

Answer:

λ =492nm

Explanation:

Given Data

diameter=2.0 cm

angular separation=3.0x10^-5 radians

λ=?

Solution

sin(theta) = 1.22 x λ /D

λ=(0.02×sin(3.0×10⁻⁵))/1.22

λ=492nm

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An internal resistance of 5 ohm and a battery of 15 ohm is connected to a resistance of 20 ohm calculate the electric current
Allushta [10]

Answer:

.6 A

Explanation:

Battery 15 VOLTS

V = IR

V / R = I

15 / ( 5+20) = .6 amps

5 0
2 years ago
In which part of the scientific method do you sum up your results
elena-14-01-66 [18.8K]
Data Analysis and Conclusion
8 0
3 years ago
A bottle rocket takes off with a = 34.5 m/s2. It
sergiy2304 [10]

Answer:3.4 seconds

Explanation:

Initial velocity(u)=0

acceleration=34.5m/s^2

Height(h)=200m

Time =t

h=u x t - (gxt^2)/2

200=0xt+(34.5xt^2)/2

200=34.5t^2/2

Cross multiply

200x2=34.5t^2

400=34.5t^2

Divide both sides by 34.5

400/34.5=34.5t^2/34.5

11.59=t^2

t^2=11.59

Take them square root of both sides

t=√(11.59)

t=3.4 seconds

8 0
3 years ago
An 8.50 m long ladder leans against the side of a building. The ladder is initially inclined at an angle of 47.0° to the horizon
erastova [34]

The angle of the ladder inclined with respect to the horizontal after being moved a distance of 0.82 m closer to the building is 53.84°

cos θ = Adjacent side / Hypotenuse

θ_{1} = 47°

Hypotenuse = Length of ladder = 8.5 m

cos 47° = Adjacent side / 8.5

Adjacent side = Initial distance of base of ladder from the building = 5.8 m

Adjacent side 2 = Final distance of base of ladder from the building

Adjacent side 2 = 5.8 - 0.82 = 4.98 m

cos θ_{2} = Adjacent side 2 / Hypotenuse

cos θ_{2} = 4.98 / 8.5 = 0.59

θ_{2} = cos^{-1} ( 0.59 )

θ_{2} = 53.84°

The formula used above is one of trigonometric ratios. Trigonometric ratios can used only in a right angled triangle where one of the angles in at 90 degrees and the other two angles are less than 90 degrees.

Therefore, the angle of the ladder inclined with respect to the horizontal after being moved is 53.84°

To know more about trigonometric ratios

brainly.com/question/1201366

#SPJ1

3 0
1 year ago
When reaching a boundary between two media (1 and 2), an incident ray is partially reflected and partially refracted. The ray is
lukranit [14]

Answer:

The angle of incidence when the reflected ray is perpendicular to the incident ray = 45°

Explanation:

According to Snell's Law,

n₁ sin θ₁ = n₂ sin θ₂

When the angle between the incident ray and reflected ray is 90°, the angle of incidence is θ₁ and the angle of reflection, θ₂ = 90° - θ₁ and the index of refraction in the Snell's Law for both media would be the same, n₁ = n₂ = n

n sin θ₁ = n sin (90° - θ₁)

Note that from trigonometric relations,

Sin (90° - θ₁) = cos θ₁

n sin θ₁ = n cos θ₁

(sin θ₁)/(cos θ₁) = 1

tan θ₁ = 1

θ₁ = arctan 1 = 45°

Hope this Helps!!!

7 0
3 years ago
Read 2 more answers
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