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puteri [66]
3 years ago
10

TIME REMISE

Physics
2 answers:
Marianna [84]3 years ago
6 0
B .It's is conserved.
SCORPION-xisa [38]3 years ago
5 0

Answer:

It is conserved

Explanation:

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What are ways to improve the design of this experiment? Check all that apply
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B,D,E

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I got you

B. Experiment with a wider range of materials.

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Thomas needs to move an 80 kg rock, but cannot lift it. He decides to use a
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Answer:

4

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3 years ago
Timed! I would really appreciate some help! thank you!
GenaCL600 [577]

Answer:

x = 5[km]

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v=x/t

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4 0
2 years ago
A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the
Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge 15\times10^-6 C is placed which is the origin.

Let B be the point where the charge 9\times 10^-6 C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, \frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2},

Q_A = 15\times 10^-6 C

Q_B=9\times10^-6C

d_A = 1+x cm

d_B=x cm

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

#SPJ4

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