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VikaD [51]
3 years ago
9

A river flows down towards a lake along an incline. initially the river is 90 m above the lake flowing at a rate of 3 m/s. at th

e lake the velocity is 0 m/s. you are interested in the change of total energy per unit mass of the water. draw a schematic of the system, clearly identifying the system, and determine the change of total energy per unit mass of the water between the inlet and outlet of the system. assume that the internal energy of the water remains constant and that the pressure at the inlet and outlet is the same.
Physics
1 answer:
lubasha [3.4K]3 years ago
4 0

Total energy =kinetic energy +potential energy

Change in energy =change in (kinetic energy +potential energy)

potential energy, P.E=mgh

where m is the mass, g is acceleration due to gravity and h is the height.

potential energy per unit mass =gh

change in potential energy per unit mass = \Delta P.E.=g\Delta h

where, h is the height.

kinetic energy= K.E. =\frac{1}{2}mv^2

change in kinetic energy per unit mass,\Delta K.E. =\frac{1}{2}\Delta v^2

In the given question:

Height varies from 90 m to zero as river flows from 90 m height to lake at 0 m

Velocity varies from 3m/s at top to o m/s at bottom.

Therefore,

\Delta E =\Delta K.E.+\Delta P.E.\\ \Rightarrow \Delta E/m=\frac{1}{2}\Delta v^2+g\Delta h=\frac{1}{2}(0^2-3^2) m^2s^{-2}+9.8 (0-9)m^2s^{-2}=(-\frac{9}{2}-88.2)m^2s^{-2}=-92.7 m^2s^{-2}

Here, it was mentioned in the question internal energy of the water is constant and there is no change in the pressure at the inlet and outlet.

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As we know that as per Newton's II law we have

F = \frac{dP}{dt}

here we will have

dP = change in momentum

dt = time interval in which momentum is changed

now in order to have least injury during jumping we need to have least force on the jumper

so in order to have least force we can say that the momentum must have to change in maximum time so that amount of force must be least

So we need to increase the time in which momentum of the system is changed

5 0
3 years ago
A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i
Hoochie [10]

The acceleration of the box is approximately 1 m/s^2

Explanation:

According to Newton's second law of motion, the net force acting on the box is equal to the product between its mass and its acceleration:

\sum F = ma

where

\sum F is the net force

m = 12.0 kg is the mass of the box

a is the acceleration

The net force can be written as

\sum F = F_a - F_f

where

F_a = 20 N is the applied forward force

F_f=9.0 N is the friction force

Combining the two equations,

F_a-F_f=ma

And solving for the acceleration,

a=\frac{F_a-F_f}{m}=\frac{20-9}{12}=0.9 m/s^2\sim 1 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

8 0
3 years ago
A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.1-m-long rope. The ball is pulled to one s
AlladinOne [14]

Answer:

The tension in the rope is 262.88 N

Explanation:

Given:

Weight W = 150 N

Length of rope r = 4.1 m

Initial speed of ball v = 5.5 \frac{m}{s}

For finding the tension in the rope,

First find the mass of rod,

mg = 150                          ( g = 9.8 \frac{m}{s^{2} } )

  m = \frac{150}{9.8}

  m = 15.3 kg

Tension in the rope is,

  T = mg + \frac{mv^{2} }{r}

  T = 150 + \frac{15.3 \times (5.5)^{2} }{4.1}

  T = 262.88 N

Therefore, the tension in the rope is 262.88 N

7 0
3 years ago
A piece of clay sits 0.10 m from the center of a potter’s wheel. If the potter spins the wheel at an angular speed of 15.5 rad/s
finlep [7]

Answer:

a=24.025\ m/s^2    

Explanation:

Given that

Distance from the center ,r= 0.1 m

The angular speed ,ω = 15.5 rad/s

We know that centripetal acceleration is given as

a=ω² r

a=Acceleration

r=Radius

ω=angular speed

a=ω² r

Now by putting the values in the above equation we get

a=15.5^2\times 0.1\ m/s^2

a=24.025\ m/s^2

Therefore the acceleration of the clay will be a=24.025\ m/s^2.

4 0
3 years ago
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