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Ainat [17]
2 years ago
14

the water behind hoover dam in nevada is 206 m higher than the colorado river below it. at what rate must water pass through the

hydraulic turbines of this dam to produce 100 mw of power if the turbines are 100 percent efficient
Physics
1 answer:
Nitella [24]2 years ago
5 0

Answer:

the required mass flow rate is 49484.37 kg/s

Explanation:

Given the data in the question;

we first determine the relation for mass flow rate of water that passes through the turbine;

so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;

W_{net} = m ( Δ P.E )

so we substitute (gh) for ( Δ P.E );

W_{net} = m (gh)

m = W_{net} / gh

so we substitute our given values into the equation

m = 100 MW / ( 9.81 m/s²) × 206 m

m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m

m = 10 × 10⁷ / 2020.86

m = 49484.37 kg/s

Therefore, the required mass flow rate is 49484.37 kg/s

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When the balloon hits the ground, it rebounds slightly. What is the source of the energy for this rebound? A. When the balloon h
nevsk [136]

Answer:

The correct answer is c.    When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL

Explanation:

Let's analyze the situation of the globe

When it touches the ground, the part that is in contact decreases its velocity to zero, but the upper part of the ball continues to move, which creates that the molecules approach slightly, if we approximate the spring links, a repulsive force is created that after all the particles reach zero speed. The force of the springs moves the ball up until the force decreases to zero.

We can relate this force of Hooke with an elastic energy

This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.

The correct answer is c

7 0
2 years ago
The power rating of an electric lawn mower is 2000 watts if lawnmower is used for 120 seconds how many joules of work van it do
Scorpion4ik [409]
1 watt = 1 joule/sec

2,000 watts = 2,000 joules/sec

                     (2,000 joule/sec) x (120 sec)

                  = (2,000 x 120)  (joule-sec/sec)

                  =   240,000 joules .
        
4 0
2 years ago
A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta
BartSMP [9]

Answer:

35%

Explanation:

Assuming no external torques present during the collision between the record and the turntable, total angular momentum must be conserved.

For a rotating body with some angular velocity and moment of inertia, the angular momentum can be expressed as follows;

L = I* ω

So, as initial angular momentum and final angular momentum must be the same, we have:

Li = Lf

⇒ I₁ * ω₁ = I₂ * ω₂ (1)

where I₁ is the rotational inertia of the turntable, and I₂, is the combined rotational inertia of the turntable and the record:

I₂ = I₁ + 0.54 I₁ = 1.54 I₁

We can solve (1) for the new common angular speed, as follows:

ω₂ = ω₁ / 1.54 (2)

The initial rotational kinetic energy is given by definition for the following equation:

Kroti = 1/2 * I₁ * ω₁² (3)

The final rotational kinetic  energy takes into account the new rotational inertia and the common final angular speed:

Krotf = 1/2* I₂ * ω₂² = 1/2* 1.54 I₁* (ω₁/1.54)² (4)

Dividing both sides in (3) and (4), we get:

Krotf/Kroti = 1/1.54 = 0.65

This means that the final rotational kinetic energy, has reduced to 0.65 of the initial value, or that has lost 35% of the initial kinetic energy.

8 0
2 years ago
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
2 years ago
Experiments using "optical tweezers" measure the elasticity of individual DNA molecules. For small enough changes in length, the
GalinKa [24]

Answer:

Spring constant, k = 0.3 N/m

Explanation:

It is given that,

Force acting on DNA molecule, F=1.5\ nN=1.5\times 10^{-9}\ N

The molecule got stretched by 5 nm, x=5\times 10^{-9}\ m

Let k is the spring constant of that DNA molecule. It can be calculated using the Hooke's law. It says that the force acting on the spring is directly proportional to the distance as :

F=-kx

k=\dfrac{F}{x}

k=\dfrac{1.5\times 10^{-9}\ N}{5\times 10^{-9}\ m}

k = 0.3 N/m

So, the spring constant of the DNA molecule is 0.3 N/m. Hence, this is the required solution.

8 0
2 years ago
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