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3 years ago

How do we use a dam to generate electricity?

Physics

2 answers:

serg [7]3 years ago

6 0

Answer:

c) Water flowing through an intake pipe causes a turbine to spin.

Explanation:

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

The energy 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.

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.

Therefore, water which is flowing through an intake pipe causes spinning of turbine which produce electricity.

umka2103 [35]3 years ago

4 0

When water from a dam passes through, turbines spin and those turbines generate electricity. This would make C the better answer.

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Some wave characteristics depend upon how the wave is produced. other wave characteristics depend upon the properties of the med

Anastaziya [24]

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<span>Mechanical waves require a medium in order to transport their energy. Sound waves are an example of mechanical way. They can not travel through vacuum. </span>

3 0

3 years ago

A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the b

aliya0001 [1]

Answer:

The torque must be applied to the wheel is 15.7 N-m.

Explanation:

Given that,

Mass of the wheel of cylinder, m = 50 kg

Diameter of the wheel, d = 1 m

Radius, r = 0.5 m

Initial speed off the wheel is 0

Final angular speed of the wheel, 120 rev/min = 12.56 rad/s

Angular displacement, $\theta=20\ rev=125.6\ rad$

The torque is given by :

$\tau=I\alpha \\\\\tau=\dfrac{mr^2}{2}\times (\dfrac{\omega_f^2}{2\theta})\\\\\tau=\dfrac{50\times1^{2}}{2}\times(\dfrac{12.56^{2}}{2\times125.6})\\\\\tau=15.7\ N-m$

So, the torque must be applied to the wheel is 15.7 N-m.

4 0

3 years ago

Match each prefix to the correct multiple of 10. 1. centi- 10 2. kilo- 100 3. deka- 1000 4. milli- 1/100 5. hecto- 1/10 6. deci-

Mice21 [21]

Explanation :

There are some basic metric conversions.

Prefix Multiple of 10

centi $\frac{1}{100}$

Kilo 1000

deka 10

milli $\frac{1}{1000}$

hecto 100

deci $\frac{1}{10}$

For example :

$1cm=\frac{1}{100}m\\\\1Km=1000m\\\\1mm=\frac{1}{1000}m\\\\1Km=10\text{ hecto meter}\\\\1Km=100\text{ deka meter}$

7 0

3 years ago

Read 2 more answers

An object has a mass of 15 kg and is accelerating to the right at 16.3 m/s2. The free-body diagram shows the horizontal forces a

marshall27 [118]

Refer to the free body diagram shown melow.

F = applied force
R = frictional force
m = 15 kg, the mass of the object

The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N

The normal reaction is
N = mg = (15 kg)*(9.8 m/s²) = 147 N
The frictional force is
R = μN = 147μ N, where μ = coefficient of kinetic friction.

Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N, (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)

Answer:
The most reasonable answer is R = 5.5 N