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klasskru [66]
1 year ago
8

The ability of a telescope to separate two closely spaced stars is called _____.

Physics
1 answer:
larisa [96]1 year ago
3 0

The ability of a telescope to separate two closely spaced stars is called (angular) resolution.

What is a telescope ?

Astronomers use a telescope to observe distant things. Curved mirrors are used by the majority of telescopes, including all large telescopes, to collect and concentrate light from the night sky. The original telescopes employed lenses, which are simply curved pieces of clear glass, to focus light.

The capacity of a telescope to distinguish two point sources into distinct pictures is known as resolution. The resolving power is constrained by diffraction effects in ideal circumstances, such as above the atmosphere where there is no turbulence (seeing).

to learn more about telescope click on the link below:

brainly.com/question/24999277

#SPJ4

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Determine the angular velocity ω of the telescope as it orbits around the Sun.
lara31 [8.8K]
The JWST is postioned about 1.5 million kilometers from the earth on the side facing away from the sun
5 0
2 years ago
A pulley with the radius of 10.0 cm was connected to a motor with a massless
kogti [31]

Answer:

(i) -556 rad/s²

(ii) 17900 revolutions

(iii) 11250 meters

(iv) -55.6 m/s²

(v) 18 seconds

Explanation:

(i) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

α = (10000 − 15000) / 9

α ≈ -556 rad/s²

(ii) Constant acceleration equation:

θ = θ₀ + ω₀ t + ½ αt²

θ = 0 + (15000) (9) + ½ (-556) (9)²

θ = 112500 radians

θ ≈ 17900 revolutions

(iii) Linear displacement equals radius times angular displacement:

s = rθ

s = (0.100 m) (112500 radians)

s = 11250 meters

(iv) Linear acceleration equals radius times angular acceleration:

a = rα

a = (0.100 m) (-556 rad/s²)

a = -55.6 m/s²

(v) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

-556 = (0 − 15000) / t

t = 27

t − 9 = 18 seconds

8 0
3 years ago
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel d
vlabodo [156]

Answer: 50π m ≈ 157 m

Explanation:

100 rev/min (2π rad/rev) / (60 sec/min) = 3⅓π rad/s

d = ωrt = 3⅓π(0.50)(30) = 50π m ≈ 157 m

8 0
3 years ago
1. A train engine pulls the boxcars. What are the horizontal forces present?(1 point)
NikAS [45]

Force is the influence that is capable of bringing about motion

The correct options are;

1. C. The pull of the engine, friction force

2. A. Weight, push of the tracks against the wheel

3. D. The astronaut weighs the most on Earth and least on the Moon

4. Please see attached drawing for example

5. Please see attached drawing for example

6. The data points will fall along a line this that as the force increases, the acceleration increases

Reason:

1. The horizontal forces are the forces acting along the horizontal x-axis, therefore;

A train engine pulling a boxcar what horizontal forces are;

C. The pull of the engine, friction force

2. The vertical forces are the forces acting an  the direction of the y-axis of a graph

The vertical forces present when a train pulls a boxcar are;

A. Weight, push of the tracks against the wheel

3. The weight of an astronaut, W = mass × Gravity

  • Therefore, the weight increases with increasing acceleration due to gravity

Therefore, the least weight is on the moon, where acceleration due to gravity is 1.6 m/s², and most on Earth where the acceleration due to gravity is approximately 9.81 m/s²

  • The correct option is D. The astronaut weighs the most on Earth and least on the Moon

4. The magnitude of the downward force is the sum of the vertical forces acting downwards

5. The net force is given by the vector sum of the forces

6. According to Newton's second law, Force, F = Mass, m × Acceleration, a

  • Therefore, where the students conduct the experiments rightly, we have that the data points will fall along a line this that as the force increases, the acceleration increases

Learn more here;;

brainly.com/question/19657850

5 0
2 years ago
Determine the direction of the effective value of g⃗ at a latitude of 45 ∘ North on the Earth.
Kitty [74]
To solve this, we simply use trigonometry
the effective value of g along the 45° angle is
g eff = g / sin 45
g eff = g / (√2 / 2)
g eff = 2g / √2
g eff = g √2 ≈ 6.94 m/s²
4 0
3 years ago
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