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3 years ago

8

Es lo mismo una energía renovable que una energía limpia?

1 answer:

Westkost [7]3 years ago

7 0

La energía limpia es energía libre de carbono que genera poca o ninguna emisión de gases de efecto invernadero. ... La energía renovable es la energía que proviene de recursos que se reponen naturalmente, como la luz solar, el viento, el agua y el calor geotérmico.

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A 60 [Hz] high-voltagepower line carries a current of 1000 [A]. The power line is at a height of 50 [m] above the earth. What is

omeli [17]

Answer:

The magnitude of the magneticfield B[T] at a point on the surface of the earth directly below the power line is B=1.496x10^(-6) T.

Explanation:

The distance from the wire to a point in the surface is the heigth of the wire.

The formula for the magnetic field on any point at distance R from a wire conducting alternating current is:

$B=\frac{\mu_0I}{2\pi R} =\frac{(4\pi\cdot 10^{-7}T\cdot m/A)(1000\,A)}{2\pi \cdot 50\,m} =1.496\cdot10^{-6}\,T$

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3 years ago

Pain in the squadron can be a sign of appemditicis

alexgriva [62]

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3 years ago

Read 2 more answers

7. What is an unbalanced force?

Step2247 [10]

Answer:

b . force that are unequal in size but in same direction.

Hope this will help . :)

8 0

3 years ago

If a crane uses 500 Joules to raise bricks 20 meters how much Newton’s of force were used

kkurt [141]

Answer:

25N

Explanation:

Given parameters:

Potential energy expended = 500J

Height = 20m

Unknown:

The force used = ?

Solution:

To solve this problem, we use the expression below:

Potential energy = Force applied x height

500 = Force applied x 20

Force applied = 25N

7 0

3 years ago

A steel ball has a diameter of

cupoosta [38]

The required increase in temperature is: $157^{\circ}C$

Explanation:

When the temperature increases, the area of an object increases according to the equation

$\Delta A = 2\alpha A_0 \Delta T$

where

$A_0$ is the initial area

$\alpha$ is the coefficient of linear expansion

$\Delta A$ is the change in area

$\Delta T$ is the change in temperature

For the steel ball in this problem, we have:

$d=0.125 m$ , so the radius is

$r=\frac{0.125}{2}=0.0625 m$

Therefore, the surface area is

$A=4\pi r^2 = 4\pi (0.0625)^2=0.0491 m^2$

The coefficient of linear expansion for steel is

$\alpha = 13\cdot 10^{-6} ^{\circ}C^{-1}$

And the change in area of the ball is

$\Delta A = 0.0002 m^2$

Therefore, the required increase in temperature is:

$\Delta T = \frac{\Delta A}{2\alpha A_0}=\frac{0.0002}{2(13\cdot 10^{-6})(0.0491)}=157^{\circ}C$

Learn more about temperature:

brainly.com/question/1603430

brainly.com/question/4370740

#LearnwithBrainly

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3 years ago