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xxMikexx [17]
3 years ago
11

Consider a river flowing toward a lake at an average velocity of 3 m/s. the river height is 90m above the lake. what is the tota

l mechanical energy of the river water per unit mass in kj/kg?
Physics
1 answer:
strojnjashka [21]3 years ago
6 0

Kinetic energy per unit of mass is

K=\frac{v^{2} }{2}

Given, v=3m/s^{2}

Therefore,

K=\frac{(3^{2} m/s^{2} )^2}{2}

K=4.5 J/kg

Now potential energy per unit mass is

p=g\times h

Given, h=90 m

Therefore,

p= 9.8m/s^2 \times 90

p=882.9 J/kg

Thus, total mechanical energy of the river water per unit mass is

T=K+p=(4.5+882.9)J/kg

T=887.9 J/kg

OR

T=0.887 kJ/kg

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A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
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Answer:

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b) The angular speed of the teenager is approximately 1.257 rad/s

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d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

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The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

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r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

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8 0
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Let it be α . Let I be moment of inertia of reel .

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