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2 years ago

2. Draw a ray diagram that shows the image

Physics

1 answer:

Lunna [17]2 years ago

6 0

you have to draw diagram

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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 $M = 109.26$

4 0

3 years ago

_ method is used to remove soluble impurities.

anygoal [31]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Methods of Separation.

Evaporation and Decantation are the process which can be easily used to remove the Soluble Impurities.

5 0

2 years ago

Read 2 more answers

What’s the most distant object in the solar system

Scrat [10]

Answer:

pluto

Explanation:

i had the same question

7 0

3 years ago

Read 2 more answers

A conducting loop has an area of 0.065 m2 and is positioned such that a uniform magnetic field is perpendicular to the plane of

aev [14]

Answer:

initial magnetic field 1.306 T

Explanation:

We have given area of the conducting loop $A=0.065m^2$

Emf induced = 1.2 volt

Initial magnetic field B = 0.3 T

Time dt = 0.087 sec

We know that induced emf is given by $e=\frac{d\Phi }{dt}=-A\frac{db}{dt}$

$1.2=0.065\times \frac{db}{0.087}$

$db=1.606T$

So initial magnetic field = 1.606-0.3= 1.306 T

5 0

2 years ago

A 1.0 kilogram ball is thrown into the air with an initial velocity of 30m/s. How much kinetic energy does the ball have? (B) ho

Gala2k [10]

Answer:

450 joules ; 450 joules ; 45.9 m

Explanation:

Given that :

Initial Velocity, u = 30m/s

Mass, m = 1 kg

Kinetic Energy of ball (KE) = 0.5mu²

K. E = 0.5 * 1 * 30^2

K.E = 0.5 * 900

K.E = 450 Joules

B.) Potential Energy (P. E)

P. E = mgh

At the highest point, all kinetic energy has would have become potential energy, hence

K. E = P. E = 450 Joules

C) Height of the ball :

From ; P. E = mgh

Where ; g = acceleration due to gravity = 9.8m/s² ; h = height

450 = 1 * 9.8 * h

450 = 9.8h

h = 450 / 9.8

h = 45.918

h = 45.9 m

7 0

3 years ago