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Kryger [21]
3 years ago
8

An object 82 cm high forms a virtual image 4.1 cm high located 4.6 cm behind a mirror. Find the object distance.

Physics
1 answer:
ioda3 years ago
8 0

Answer:

The object distance is 92 cm.  

Explanation:

let v be the image distance and h be the height of the image, let u the be the object distance and H be the height of the object.

then, the magification of the mirror is given by:

m = -v/u and m = h/H

so, -v/u = h/H

         u = -v×H/h

            = -(-4.6)×(82)/(4.1)

            = 92 cm

Therefore, the object distance is 92 cm.

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a rocket, initially at rest, is fired vertically with a net upward acceleration of 12 m/s2 . at an altitude of 0.50 km, the engi
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The rocket travelled a maximum height at 1.0102 km.

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The acceleration of a rocket (a) = 12 m/s²

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A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.

The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²

(a) = Δv/Δt

Where , Δv is change in velocity and Δt is change in time.

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Where,Δx is the change in position and Δt is change in time & v is velocity.

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v² = 0 + ( 2× 10 × 0.5×10³)m/s

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v = 100 m/s

Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,

v² = u² - 2(g)h

h = (v²- u² ) / 2g

h = 10,000/2×9.8

h = 510.2 m

So that the rocket travelled the maximum height ,

(h)= (0.5 km + 510.2m)

(h) = 1.0102 km

Hence, the rocket travelled at the maximum height h is 1.0102 km

To know more about acceleration

brainly.com/question/15135960

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