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2 years ago

A 5 kg rock falls from an 8 m high cliff. What is the speed of the rock when it hits the ground?

Physics

1 answer:

Levart [38]2 years ago

3 0

Answer:

3k mph

Explanation:

don't take my answer it is wrong

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If you decrease the length of a wire, what effect will it have on resistance

attashe74 [19]

As the length increases, resistance increases, as a result current decreases.

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3 years ago

1. why is<br> experimentation so important to science

Alekssandra [29.7K]

Answer: to provide evidence to a theory

Explanation: Experimentation allows for multiple trials to provide evidence to a scientific theory. Without experimentation there would be no data to back up your hypothesis.

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3 years ago

*PLZ HELP ASAP* Saturn and Jupiter both complete one rotation in about____hours.

Debora [2.8K]

Answer:

Explanation: Earth takes 24 hours to complete one spin, and Mars takes 25 hours. The gas giants rotate really fast. Jupiter takes just 10 hours to complete one rotation. Saturn takes 11 hours, Uranus takes 17 hours, and Neptune takes 16 hours.

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2 years ago

Read 2 more answers

What is the greatest distance an image can be located behind a convex spherical mirror?

Afina-wow [57]

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

$\frac{1}{d_i} + \frac{1}{-d_o} = \frac{1}{f}$

$\frac{1}{d_i} = \frac{1}{d_o} + \frac{1}{f}$

so here maximum value of image distance is equal to focal length of the mirror

6 0

2 years ago

A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge

Kisachek [45]

Answer:

$\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

$\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r$

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

$\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

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3 years ago