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3 years ago

How does hydroelectric energy works?

1 answer:

7 0

Actually, hydroelectric and coal-fired power plants produce electricity in a similar way. In both cases a power source is used to turn a propeller-like piece called a turbine, which then turns a metal shaft in an electric generator,which is the motor that produces electricity. A coal-fired power plant uses steam to turn the turbine blades; whereas a hydroelectric plant uses falling water to turn the turbine. The results are the same. :) <span>
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A skydiver jumps from an airplane and falls freely for 5 seconds. If the sky diver's velocity increases by 49 m/s during that ti

Sergeeva-Olga [200]

Acceleration = (change in speed) / (time for the change)

   =    (49 m/s) / (5 seconds)

   = (49 / 5) m/s / s

   =    9.8 m/s²

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3 years ago

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The colour of star depend on its temperature, why?

taurus [48]

<em>Another key factor that determines a star's colour is its temperature. As stars become hotter, the overall radiated energy increases, and the peak of the curve changes to shorter wavelengths. To put it another way, when a star heats up, the light it produces moves toward the blue end of the spectrum.</em>

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3 years ago

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener

defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

$\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s$

(a) the force constant of the spring

$\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m$

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

$A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m$

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3 years ago

Can someone please help me ASAP???!!!

olganol [36]

Answer: The answers are:

Explanation:

1. C

2. A

3. C

4. B

5. D

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2 years ago

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3 years ago