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3 years ago

14

The learning about Universe is called ________ ...

2 answers:

ASHA 777 [7]3 years ago

7 0

Answer:

The learning about Universe is called

<u>Astronomy</u><u>.</u>

Explanation:

<h2>Hope it Helps you!! </h2>

FrozenT [24]3 years ago

5 0

The learning about Universe is

called <u>Astronomy</u>...

Astronomy - the branch of science which deals with celestial objects, space, and the physical universe as a whole.

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Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the

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Explanation:

Given

mass of Susan $m=12 kg$

Inclination $\theta =30^{\circ}$

Tension $T=29 N$

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Resolving Forces Along x axis

$F_x=T\cos \theta -f_r$

where $f_r=friction\ Force$

$F_y=mg-N-T\sin \theta$

since there is no movement in Y direction therefore

$N=mg-T\sin \theta$

and $f_r=\mu N$

Thus $F_x=T\cos \theta -\mu N$

$F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))$

$F_x=25.114-18.558$

$F_x=6.556 N$

Work done by applied Force is equal to change to kinetic Energy

$F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2$

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$v_f^2=\frac{6.556\times 2.7\times 2}{12}$

$v_f^2=2.95$

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

notka56 [123]

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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