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3 years ago

This type of liquid can be used as an electric tester Is that a good conducted of electricity?

Physics

1 answer:

Helga [31]3 years ago

7 0

Answer:

Yes

Explanation:

Testing electricity is a good conductor of electricity.

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If the bar magnet is flipped over and the south pole is brought near the hanging ball, the ball be 1) I'am not sure 2) Unaffecte

igomit [66]

If the ball is made of any magnetic metals (Iron, Nickel, Cobalt) It's 3.

But if it's made of something else without any magnetic force it's 2.

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3 years ago

What was a pattern in nature that contributed to an understanding of atoms?

devlian [24]

the fact that chemicals have different colors

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2 years ago

20 points, just need a basic understanding. You are on a mystery planet, what you know is that from a height of 10.0 meters, a d

ICE Princess25 [194]

(since you asked for basic understanding only, I am not including actual calculations. Please let me know in the comments section if you wish to verify your solution(s))

For (b): Use the formula for distance (s) made during an accelerated motion:

$s = \frac{1}{2}at^2+ v_0t+s_0= \frac{1}{2}at^2= \frac{1}{2}gt^2$

with v_0 and s_0 being the initial velocity and distance, both 0 in this case, and with "a" denoting the acceleration, in this case solely due to gravitational acceleration so: "g."

You are given the distance made, namely 10 m, and the duration t (0.88s) and so using the formula above you can solve for g.

For (c), to determine the final velocity at time 0.88s use the formula for the instantaneous velocity of an accelerated motion

(velocity at time t) = (acceleration) x (time)

again, with acceleration due to gravity, i.e., a = g and with g as determined under (b).

If my calculation is correct, this mystery planet could be the Jupiter.

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4 years ago

The speed at which a wave is transmitted depends on _____. A. the state of matter of the medium B. whether or not the wave is re

11Alexandr11 [23.1K]

Answer: C

Explanation:

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3 years ago

The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec

Elanso [62]

On Earth, the period of a pendulum is given by:
$T_{earth}=2\pi \sqrt{ \frac{L}{g_{earth} }$
where L is the length of the pendulum and $g_{earth}=9.81~m/s^2$ is the gravitational acceleration on Earth.
Similarly, the period of the same pendulum on Mars will be
$T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }$
where $g_{mars}=3.71~m/s^2$ is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
$\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} } = \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63$
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.

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4 years ago