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3 years ago

15

"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e

yepiece whose focal length is 2.7 cm

1 answer:

notka56 [123]3 years ago

4 0

Answer:

The Magnifying power of a telescope is $M = 109.26$

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective $f_{objective}$ = $\frac{R}{2}$

⇒ $f_{objective}$ = $\frac{590}{2}$

⇒ $f_{objective}$ = 295 cm

Focal length of eyepiece $f_{eyepiece}$ = 2.7 cm

Magnifying power of a telescope is given by,

$M = \frac{f_{objective} }{f_{eyepiece} }$

$M = \frac{295}{2.7}$

$M = 109.26$

therefore the Magnifying power of a telescope is $M = 109.26$

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Answer:

The final velocity of the block is 68.85m/s.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

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$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

But it is necessary to know the acceleration. For a better procedure it will be listed the knowns and unknowns of the problem:

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m = 1.75 Kg

d = 23.9 m

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Unknowns:

$F_{r}$ = ?

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The acceleration can be found by means of Newton's second law:

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Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

<em>Force in the x axis:</em>

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<em>Forces in the y axis: </em>

$F_{y} = N - W_{y}$ (4)

<em>Solving for the forces in the x axis:</em>

$F_{x} = F + W_{x} - F_{r}$

Notice that is necessary to found $F_{r}$ :

$F_{r} = \mu N$ (5)

The normal force can be obtained from equation (4)

$N - W_{y} = 0$

$N = W_{y}$

The component of the weight in the y axis can be gotten by means of trigonometry:

$\frac{Adjacent}{Hypotenuse} = cos \theta$

$\frac{W_{y}}{W} = cos \theta$

$W_{y}= W cos \theta$

Remember that the weight is defined as:

$W = mg$

$W_{y}= mgcos \theta$

$N = mg cos \theta$

$F_{r} = \mu mgcos \theta$

$F_{r} = (0.136)(1.75Kg)(9.8m/s^{2})(cos30)$

$F_{r} = 2.01N$

The component of the weight in the x axis can be gotten by means of trigonometry:

$\frac{Opposite}{Hypotenuse} = sen \theta$

$\frac{W_{x}}{W} = sen \theta$

$W_{x} = W sen \theta$

$W_{x} = mgsen \theta$

$W_{x} = (1.75Kg)(9.8m/s^2)(sen30)$

$W_{x} = 8.57N$

Then, replacing $W_{x}$ and $F_{r}$ in equation (3) it is gotten:

$F_{x} = 167N + 8.57N - 2.01N$

$F_{x} = 173.56N$

Solving for the forces in the y axis:

$F_{y} = N - W_{y}$

$F_{y} = mgcos \theta - mgcos \theta$

$F_{y} = 0$

Replacing the values of $F_{x}$ and $F_{x}$ in equation (2) it is gotten:

$F_{x} + 0 = ma$

$F_{x} = ma$

$a = \frac{F_{x}}{m}$

$a = \frac{173.56N}{1.75Kg}$

$a = \frac{173.56Kg.m/s^{2}}{1.75Kg}$

$a = 99.17m/s^{2}$

Now that the acceleration is known, equation (1) can be used:

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$

However, since the block was originally at rest its initial velocity will be zero ( $v_{i} = 0$ ).

$v_{f} = \sqrt{2ad}$

$v_{f} = \sqrt{2(99.17m/s^{2})(23.9m)}$

$v_{f} = \sqrt{2(99.17m/s^{2})(23.9m)}$

$v_{f} = 68.85m/s$

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