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ryzh [129]
3 years ago
12

A 64 kg swimmer jumps, with a velocity of 4.2 m/s, off the front of a 25 kg kayak when the kayak is moving forward at a velocity

of 3.2 m/s. What is the velocity of the kayak after the swimmer jumps off
Physics
1 answer:
Crank3 years ago
8 0

Answer:

3.88m/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and 2 are the initial velocities

v is the final velocity

Given

m1 = 64kg

u1 = 4.2m/s

m2 = 25kg

u2 = 3.2m/s

Required

Final velocity v

Substitute the given values into the formula

64(4.2)+25(3.2) = (65+25)v

268.8+80 = 90v

348.8 = 90v

v = 348.8/90

v = 3.88m/s

Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s

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AnnyKZ [126]

The work done by the turbine will be 708.2 kJ/kg. The work done by the turbine is the difference of the enthalpy at inlet and exit.

<h3 /><h3>What is temperature?</h3>

Temperature directs the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles, more the temperature.

If the given turbine is assumed to be reversible;

\rm P_I(Initial pressure)=60 mpa = 60 bar

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The change in the entropy is;

\rm S_2-S_1=\frac{\delta q}{dt}

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How does the current in a circuit change if the resistance is doubled? The current is halved. The current is doubled. The curren
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Answer:

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Explanation:

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Answer:

Explanation:

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