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nata0808 [166]
3 years ago
10

Which proportion can you use to find the value of a? (posted in this category because of bots)

Physics
1 answer:
pantera1 [17]3 years ago
5 0

Answer:

But you a BjTCH

Explanation:

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The stage where atoms are spread out and bouncy is the gas stage.

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A point charge is placed 100 m from a 6 uC charge generating an electric field. What is the
kvv77 [185]

The strength of the electric field is 5 N/C

Explanation:

The magnitude of the electric field produced by a single-point charge is given by:

E=\frac{kQ}{r^2}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

Q=6 \mu C = 6\cdot 10^{-6}C is the charge producing the field

r = 100 m is the distance from the charge at  which we want to calculate the field

Substituting into the equation, we find the s trength of the electric field:

E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C

Learn more about electric field:

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3 0
3 years ago
R'=2 Ohm R"=1,5 Ohm R"'=2 Ohm​
Andrew [12]

Answer:

2.5 ohm

Explanation:

R' and R''' are parallel

So,

1/R1= 1/R' + 1/R'''

1/R1 = 1/2 + 1/2

1/R1 = 1

so,

R1= 1 ohm

Now R1 and R'' are in series

so,

R= R1 + R''

R= 1 + 1.5

R= 2.5 ohm

5 0
3 years ago
A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?
garri49 [273]
<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>
6 0
3 years ago
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