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attashe74 [19]
3 years ago
14

Hoover Dam on the Colorado River is the highest dam in the United States at 221m, with a power output of 680 MW. The dam generat

es electricity by flowing water down to a point 150 m below the stop, at an average flow rate of 650 m3/s.
Part (a) Calculate the power in this flow in watts.
Physics
2 answers:
Lapatulllka [165]3 years ago
8 0

Answer:

power = 1407.77 MW

Explanation:

The basic principle of a hydro electric station is the conversion of potential energy to electrical energy. Here, water is allowed to fall from a height which will increase its the kinetic energy. This high speed flowing water is used to rotate the shaft of a turbine which will in turn produce electrical energy.

So here,

power = rate of change of potential energy with respect to time

power = \frac{d(mgh)}{dt}

where,

m = mass of water

g = acceleration due to gravity

h = height through which the water falls

here h and g are constants ( h is the total height through which the water falls and it doesn't change with time). Therefore we can take them out of differentiation.

thus,

power = gh\frac{d(m)}{dt}

now,

m = ρV

where,

ρ = density

V = volume

substituting this in the above equation we get,

power = gh\frac{d(ρV)}{dt}

again ρ is a constant. Thus,

power = ρgh\frac{d(V)}{dt}

Given that,

h = 221 m

\frac{d(V)}{dt} = 650 m^{3}/s

g = 9.8 m/s^{2}

ρ = 1000 kg/m^{3}

substituting these values in the above equation

power =  1000 x 9.8 x 221 x 650

power = 1407.77 MW

Phantasy [73]3 years ago
4 0

Answer:

<u> Power = 9.75 ×10^8\frac{kgm^2}{s^3}</u>

Explanation:

  • Power is rate of change of energy.
  • Here gravitational energy is transferred to kinetic energy of water at a definite rate.

For one second 650m^3 of water flows out down to 150m oh depth.

So, the energy at a height of 150m is transformed to kinetic energy.

for a second,

       650m^3 of water flows down ⇒ (1000kg/m^3 × 650m^3) = 6.5×10^5kg of warer flos down.

The total gravitational potential energy stored in water is

    = <u>mass of water × height× gravity</u>

    = 6.5 ×10^5 × 150 × 10 =  9.75 ×10^8\frac{kgm^2}{s^2}

As it is transformed in a second it is also equal to <u>Power.</u>

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Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

U=(M+m)gh       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

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h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m

The potential energy is:

U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

Mv_1+mv_2=(M+m)v    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

M(0)+mv_2=(M+m)v    

v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N    

The force on the cord after the impact is 2.59N

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