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olga_2 [115]
3 years ago
8

A hydraulic turbine is used to generate power by using the water in a dam. The elevation difference between the free surfaces up

stream and downstream of the dam is 131 m. The water is supplied to the turbine at a rate of 201 kg/s. If the shaft power output from the turbine is 234 kW, the efficiency of the turbine is_________.
Physics
1 answer:
Serjik [45]3 years ago
3 0

Answer:

0.906

Explanation:

Let g = 9.81 m/s2. We can calculate the rate of change in potential energy when m = 201kg of water is falling down a distance of h = 131m per second

\dot{E_p} = \dot{m}gh = 201*9.81*131 = 258307 J/s (W) = 258.307 kW

So the efficiency of the water turbine is the ratio of output power over input power:

\frac{234}{258.307} = 0.906

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The tent meteorite, found in 1897 near cape York, on the west coast of Greenland, is the largest meteorite exhibited by any muse
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3 years ago
The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about
nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

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3 years ago
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6 0
2 years ago
Read 2 more answers
A straight segment of a current-carrying wire has a current element IL where I = 2.70 A and L = 3.20 cm i + 4.30 cm j. The segme
myrzilka [38]

The component of the force in negative z-direction is -0.144 N.

The given parameters;

  • <em>current in the wire, I = 2.7 A</em>
  • <em>length of the wire, L = (3.2 i + 4.3j) cm</em>
  • <em>magnetic filed, B = 1.24 i</em>

The force on the segment of the wire is calculated as follows;

F = ILBsin(\theta)

where;

  • <em>θ is the angle wire and magnetic field</em>

<em />

The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0

The magnitude of the wire length is calculated as follows;

|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m

The component of the force in negative z-direction is calculated as;

F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times  sin(53.3)\\\\F_z = -0.144 \ N

Thus, the component of the force in negative z-direction is -0.144 N.

Learn more here:brainly.com/question/22719779

6 0
2 years ago
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