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myrzilka [38]
3 years ago
5

A 5 kg fish swimming 2 m/s swallows an absent minded 1 kg fish swimming toward it at a velocity that brings both fish to a halt

immediately after lunch. What is the velocity v of the smaller fish before lunch?
Physics
1 answer:
Gelneren [198K]3 years ago
8 0

Answer:

v = 10 m/s

Explanation:

given,

Mass of large fish, M = 5 Kg

speed of swimming, v' = 2 m/s

mass of small fish, m = 1 kg

speed of the fish = v

using conservation of momentum

m v + M v' = M u' + m u

final speed of both the speed is zero.

- 1 x v + 5 x 2  = 0

negative sign is used because small fish is moving in opposite direction.

now,

v = 10 m/s

hence, the speed of the small fish is equal to 10 m/s.

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Answer:

a) i = -9.63 cm ,    h ’= .0.24075 cm   erect

b)  i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

               1 / f = (n-1) (1 / R₁ - 1 / R₂)

                1 / f = (1.70 -1)) 1 / ∞ - 1/13)

                1 / f = 0.0538

                 f = - 18.57 cm

Now we can use the constructor equation

             1 / f = 1 / o + 1 / i

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               1 / i = -0.1038 cm

               I = -9.63 cm

For the height of the

image let's use magnification

                 m = h '/ h = - i / o

                  h ’= -h i / o

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                  h ’= .0.24075 cm

b) we invert the lens

The focal length is

             1 / f = (1.70 -1) (1/13 - 1 / int)

              1 / f = 0.0538

             f = 18.57 cm

             1 / i = 1 / f -1 / o

             1 / I = 1 / 18.57 - 1/20

             1 / I = 3.85 10-3

             i = 259.74 cm

     

            h ’= - 0.5 259.74 / 20

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3 years ago
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Gay-Lussac's Law states
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I don't think we know enough about the experimental set-up and methods
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