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2 years ago

Is electricity a fuel

2 answers:

8 0

YES, ELECTRICITY CONCERNS ENERGY WHICH IS USED AS A FUEL . IN MODERN DAY TECH, MOST MACHINES USE ELECTRICITY AS A FUEL SUCH AS THE ELECTRONIC TRAIN IN TOKYO, JAPAN.

Anestetic [448]2 years ago

3 0

Answer:

Electricity is considered an alternative fuel under the Energy Policy Act of 1992. Electricity can be produced from a variety of energy sources, including natural gas, coal, nuclear energy, wind energy, hydropower, as well as solar energy and stored as hydrogen or in batteries.

Explanation:

hope this helps

brainliest plz

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PLz hlp me!!!! ASAP Will mark brainliest

Troyanec [42]

Answer:

hurricane

Explanation:

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3 years ago

Read 2 more answers

What is true of an object pulled inward in an electric field?

slava [35]

Answer:

option b

Explanation:

There is an object pulled inward in an electric field.

We have to find out of the four options given which is true.

a) The object has a neutral charge is false since when electric field pulls the object inward, there is a charge inside.

b) The object has a charge opposite that of the field, this option is correct since there will be an equal and opposite charge created by the object

c) The object has a negative charge will be correct only if the original charge was positive hence wrong

d) The object has a charge the same as that of the field is incorrect since this would be opposite the charge

So only option b is right

5 0

3 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

2 years ago

Someone please help me

Schach [20]

Answer:

300x480 teaspoons

Explanation:

when converting cups to teaspoons just multiply by 48

3 0

3 years ago

Read 2 more answers

What occurs when water droplets in the atmosphere become heavier than the air surrounding them can hold?