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erma4kov [3.2K]
3 years ago
11

A 5 kg fish swimming at 1 m/s swallows an absentminded 500 g fish swimming toward it at a velocity that brings both fish to a ha

lt immediately after lunch. What is the velocity of the approaching smaller fish before lunch?
Physics
1 answer:
AlexFokin [52]3 years ago
8 0

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum is defined as the product between mass and velocity of each body. And its conservation as the equality between the initial and final momentum. Mathematically described as

m_1u_1+m_2u_2 = (m_1+m_2)v_f

Here

m_1 = Mass of big fish

m_2 = Mass of small fish

v_1 = Velocity of big fish

v_2 = Velocity of small fish

v_F = Final Velocity

The big fish eats small fish and the final velocity is zero. Rearrange the equation for the initial velocity of small fish we have

m_1u_1=-m_2u_2

u_2 = -\frac{m_1u_1}{m_2}

Replacing we have,

u_2 = -\frac{(5kg)(1m/s)}{0.5kg}

u_2 = -10m/s

The negative sign indicates that the small fish is swimming in the direction opposite to that of the big fish.

Therefore the speed of the small fish is 10m/s

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2 years ago
A daring 510-N swimmer dives off a cliff with a running horizontal leap. What must her minimum speed be just as she leaves the t
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Answer:

v_x = 1.26 m/s

Explanation:

given,

weight of swimmer = 510 N

length of ledge, L = 1.75 m

vertical height of the cliff, h =  9 m

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1.75 = v_x × t ........(1)

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1.75 = v_x × 1.39

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3 years ago
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Answer:

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Explanation:

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v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

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