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PilotLPTM [1.2K]
3 years ago
5

To form a better telescope, you can interchange eyepieces for new focal lengths. A new design lets you replace both lenses, whic

h new configuration would make a more powerful telescope?
a. a short focal length objective and a short focal length eyepiece.
b. a long focal length objective and a short focal length eyepiece.
c. a long focal length objective and a long focal length eyepiece.
d. a short focal length objective and a long focal length eyepiece.
Physics
1 answer:
HACTEHA [7]3 years ago
8 0

Answer:

option (b)

Explanation:

for a telescope, the object is placed very far from the objective lens so the focal length of objective is large and the eye piece is of short focal length.

So, long focal length objective and short focal length eyepiece.

Option (b) is correct.

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Kyle has a mass of 54 kg and is jogging at a velocity of 3 m/s. What is Kyle’s kinetic energy?
olga nikolaevna [1]

Answer:243joules

Explanation:

Mass(m)=54kg

Velocity(v)=3m/s

Kinetic energy =(m x v^2)/2

Kinetic energy =(54 x 3^2)/2

Kinetic energy =(54 x 9)/2

Kinetic energy =486/2

Kinetic energy =243joules

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4 years ago
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Explain why a greater frequency indicates greater energy in a wave.
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<span> Given the relationship between </span>wavelength<span> and </span>frequency<span> — the </span>higher<span>the </span>frequency<span>, the shorter the </span>wavelength<span> — it follows that short wavelengths are</span>more<span> energetic than long wavelengths.</span>
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3 years ago
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A force of 6.0 N gives a 2.0 kg block an acceleration of 3.0
Semenov [28]

Explanation:

The Net Force of the object can be written by:

Fnet = ma

where m is the mass of the object in <em>kg</em>

a is the acceleration of the object in <em>m/s^2</em>

Hence by applying the formula we get:

Fnet = (2.0)(3.0)

= 6N

We also know that Net force is also the sum of all forces acting on an object. In this case Friction and the Pushing Force is acting on the object. Hence we can write that:

Fnet = Pushing Force + (-Friction)

6N = 6N - Friction

Friction = 0N

Hence the<u> </u><u>f</u><u>orce of friction is 0N.</u>

7 0
3 years ago
A quarterback back pedals 3.3 meters southward and then runs 5.7 meters northward. For this motion, what is the distance moved?
mixas84 [53]

Answer:

The distance moved is 9 meters

The magnitude of the displacement is 2.4 meters

The direction of the displacement is northward

Explanation:

- <em><u>Distance</u></em> is the length of the actual path between the initial and the

  final position. Distance is a scalar quantity

- <u><em>Displacement</em></u> is the change in position, measuring from its starting

  position to the final position. Displacement is a vector quantity

The quarterback pedals 3.3 meters southward

That means it moves down 3.3 meters

Then runs 5.7 meters northward

That means it runs up 5.7 meters

The distance = 3.3 + 5.7 = 9 meters

<em></em>

<em>The distance moved is 9 meters</em>

It moves southward (down) for 3.3 meters and then moves northward

(up) for 5.7

It moves from zero to 3.3 down and then moves up to 5.7

The displacement = 5.7 - 3.3 = 2.4 meters northward

<em />

<em>The magnitude of the displacement is 2.4 meters</em>

<em>The direction of the displacement is northward</em>

8 0
3 years ago
In the parts that follow select whether the number presented in statement A is greater than, less than, or equal to the number p
Brums [2.3K]

Answer:

a) Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

Explanation:

a) Statement A : 2.567km to two significant figures.

2.567km 2. S.F = 2.6km

Statement B : 2.567km to three significant figures.

2.567km 3 S.F = 2.57km

Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) statement A: (2.567 km + 3.146km) to 2 S.F

(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km

Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).

2.567km to 2 S.F = 2.6km

3.146km to 2 S.F = 3.1km

2.6km + 3.1km = 5.7km

Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

5 0
3 years ago
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