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2 years ago

What two factors determine the energy production of the Hoover Dam

2 answers:

swat322 years ago

6 0

Mar 28, 2011 · The Hoover Dam generates electricity using hydropower. Hydropower is the power generated by moving water. Dams are often built on a large river, so the water falls down from the top of the dam. At the bottom of the dam, the water rotates a large turbine

Hope this helps :p

scZoUnD [109]2 years ago

4 0

Answer:

Two factors are water speed and amount of water.

Explanation:

The production of hydro electrical power can be done when the water passes from the dam to the rivers which are below dam.

Turbines which are surrounded by magnets produce the electricity when the flowing water is hit on the turbines than turbines rotates and produce electricity and turbines are present in the dam.

The energy production of the Hoover dam is based on the amount or pressure of the water if the more water is passing through the intake pipe then there is more production of energy. An artificial lake is created behind the dam.

And energy production also depends on the speed of water means if the speed of water is large then it rotate the turbine faster and increase the production.

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In the following equations, the distance x is in meters, the time tin seconds. and the velocity v is in meters per second. whata

lyudmila [28]

Answer:

A) c₁ = m, c₂ = m/s

B) c₁ = m/s²

C) c₁ = m/s²

D) c₁ = m/s c₂ = °

E) c₁ = m/s , c₂ = /s

Explanation:

A) x = c₁ + c₂t

⇒m = m + (m/s)s (Only same units can be added)

⇒m = m

So, c₁ = m, c₂ = m/s

B) x = 0.5c₁t²

⇒m = 0.5 (m/s²)s²

⇒m = m

So, c₁ = m/s²

C) v² = 2c₁x

⇒m²/s² = 2 (m/s²)m

⇒m²/s² = m²/s²

So, c₁ = m/s²

D) x = c₁ cos(c₂)t

⇒m = (m/s) cos(°)s

⇒m = m

So, c₁ = m/s c₂ = °

E) v² = 2c₁v-(c₂x)²

⇒m²/s² = 2(m/s)(m/s)-(1/s²)(m²)

⇒m²/s² =m²/s²

So, c₁ = m/s , c₂ = /s

3 0

1 year ago

Read 2 more answers

How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.

Rasek [7]

Answer:

t = 4.17 [s]

Explanation:

We know that work is defined as the product of force by distance.

W = F*d

where:

F = force [N] (units of Newtons)

d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]

In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.

w = m*g

where:

m = mass = 1.47 x 10⁴ [g] = 14.7 [kg]

g = gravity acceleration = 9.81 [m/s²]

w = 14.7*9.81

w = 144.2 [N]

Therefore the work can be calculated.

W = w*d

W = 144.2*63.4

W = 9142.72 [J] (units of Joules)

Power is now defined in physics as the relationship of work at a given time

P = W/t

where:

P = power = 2190 [W]

t = time [s]

Now clearing t, we have.

t = W/P

t = 9142.72/2190

t = 4.17 [s]

6 0

1 year ago

A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s

sweet [91]

This question is in complete.The question is

A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.

Answer:

distance=0.124 m

Explanation:

$mgh=mglSin\alpha =(1/2)Iw_{i}^{2}+(1/2)mv^{2}\\ v=wR\\Solve for L\\L=((1/2)(1/2)0.015^{2}*60^{2}+(1/2)(60*0.015^{2} ))/9.8Sin30\\ L=0.124m$

6 0

2 years ago

An automobile traveling at 25.0 km/h along a straight, level road accelerates to 65.0 km/h in 6.00 s. what is the magnitude of t

USPshnik [31]

Note that
1 km/h = (1000 m)/(3600 s) = 0.27778 m/s

Initial velocity, v₁ = 25 km/h = 6.9444 m/s
Final velocity, v₂ = 65 km/h = 18.0556 m/s

Time interval, dt = 6 s.

Calculate average acceleration.
a = (v₂ - v₁)/dt
= (18.0556 - 6.9444 m/s)/(6 s)
= 1.852 m/s²

Answer:
The average acceleration is 1.85 m/s² (nearest hundredth)

3 0

1 year ago

Two chargedparticles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged part

erica [24]

Answer:

Two possible points

x= 0.67 cm to the right of q1

x= 2 cm to the left of q1

Explanation:

Electrostatic Forces

If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

$\displaystyle f=k\frac{q_1\ q_2}{d^2}$

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.

$\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}$

$\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}$

$\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}$

$\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}$

$\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}$

Equating

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}$

Operating and simplifying

$\displaystyle (0.02-x)^2=4x^2$

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

$\displaystyle 0.02-x=\pm 2x$