Physics, asked by kvngfx9586, 9 hours ago

A bird in air is diving vertically over a tank with speed 5 cm/s. Base of tank is silvered. A fish in the tank is rising upward along the same line with speed 2 cm/s. Water level is falling at rate of 2 cm/s.

Answers

Answered by sn2nidhi
0

Explanation:

8↑w.r.tfish=8-4=4↑(A) speed of fish seen from air

v

fish

air

=v

fish

water

v

fish

air

=4/

3

4

v

fish

air

=3 upwards

=4×34=3↑

speed of fish w.r.t bird

v

fish

bird

=v

fishair

+v

bird

v

fish

bird

=3+6

v

fish

bird

=9 upwards

8↑w.r.tfish=8-4=4↑(B) Speed of the image of fish formed after reflection from the mirror w.r.t air

v

fish

air

=v

fish

water

v

fish

air

=4/

3

4

v

fish

air

=3

speed of fish w.r.t bird

v

fish

bird

=−v

fishair

+v

bird

v

fish

bird

=−3+6

v

fish

bird

=3

8↑w.r.tfish=8-4=4↑(C) speed of bird seen from water

v

bird

water

=v

bird

water

v

bird

water

=6.

3

4

v

bird

water

=8 downwards

=4×34=3↑

speed of bird seen w.r.t fish

v

bird

fish

=v

birdwater

+v

fish

v

bird

fish

=8+4

v

bird

fish

=12 downwards

8↑w.r.tfish=8-4=4↑(D) Speed of image of bird w.r.t water looking downward in the mirror

v

bird

water

=v

bird

water

v

bird

water

=6.

3

4

v

bird

water

=8

=4×34=3↑

w.r.t fish

v

bird

fish

=v

birdwater

−v

fish

v

bird

fish

=8−4

v

bird

fish

=4

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