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guajiro [1.7K]
3 years ago
6

What is electropower​

Physics
2 answers:
OLEGan [10]3 years ago
8 0

Answer:

electricity

Explanation:

Sloan [31]3 years ago
7 0

<em>because its the same thing as lights </em>

Explanation:

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What particles bond together to
yanalaym [24]

Answer:

atoms bond together and forms molecules

5 0
2 years ago
Read 2 more answers
Name a situation when potential energy is useful
adell [148]
Gravitational energy is a form of potential energy because it is dependent on the mass of an object and needs to be calculated for the specific object.
3 0
3 years ago
A 12.0-g plastic ball is dropped from a height of 2.50 m. Just as it strikes the floor, it is moving at a speed of 3.20 m/s. How
nalin [4]

Answer:

0·233 J

Explanation:

Given

Mass of the ball = 0·012 kg

Initially the ball is at a height of 2·5 m

As initially the ball is dropped, it's initial velocity will be equal to 0

Therefore initially it has zero kinetic energy and has only potential energy

∴ Initially total mechanical energy of the ball = potential energy of the ball

Initial potential energy of the ball = m × g × h

where

m is the mass of the ball

g is the acceleration due to gravity

h is the height of the ball

∴ Potential energy = 0·012 × 9·8 × 2·5 = 0·294 J

Velocity of the ball after striking the floor = 3·2 m/s

After striking the floor, the total mechanical energy = kinetic energy just after striking the floor

Kinetic energy = 0·5 × m × v²

where m is the mass of the ball

v is the velocity of the ball

∴ Kinetic energy of the ball = 0·5 × 0·012 × 3·2² = 0·061 J

Mechanical energy that is lost = 0·294 - 0·061 = 0·233 J

∴ Mechanical energy that the ball lost during its fall = 0·233 J

6 0
2 years ago
Help in my home work!<br>convert 80°c to fahrenheit?​
Vsevolod [243]

Answer:

176 degrees fahrenheit

Explanation:

80°C × 9/5) + 32 = 176°F

Mark me brainliest!!

7 0
2 years ago
a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column
scoray [572]

Answer:

The answer is "1155\ \frac{kg}{m^3}"

Explanation:

Please find the complete question in the attached file.

p = p_0 + ?gh

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

h_1 = 11 \ cm\\\\h_2= 3 \ cm

The pressures would be proportional to the quantity 11-3 = 8 cm from below the surface at the interface between both the oil and the liquid.

\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\

       = \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}

8 0
2 years ago
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