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Liono4ka [1.6K]
3 years ago
9

NEED HELP Two skaters stand facing each other. One skater’s mass is 60 kg, and the other’s mass is 72 kg. If the skaters push aw

ay from each other without spinning,Single choice.
(2 Points)

the lighter skater has less momentum.

the heavier skater has less momentum

their total momentum decreases.

their momenta are equal but opposite.
Physics
2 answers:
Masja [62]3 years ago
4 0

Answer:

Two skaters stand facing each other. One skater's mass is 60 kg, and the other's mass is 72 kg. If the skaters push away from each other without spinning. Their momentums are equal but opposite.

Explanation:

Answer option D) Their momentums are equal but opposite.

stealth61 [152]3 years ago
3 0

Answer:

the heavier skater has less momentum

hope it is helpful to you

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Answer:

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But, ΔU, the change in internal energy is independent of pathway

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A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 cm , giving it a ch
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A) The electric field inside the paint layer is zero

B) The electric field just outside the paint layer is 3.2\cdot 10^7 N/C (radially inward)

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Explanation:

A)

We can solve the problem by applying Gauss Law, which states  that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:

\int EdS = \frac{q}{\epsilon_0}

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\epsilon_0 is the vacuum permittivity

By taking a sphere centered in the origin,

\int E dS = E \cdot 4\pi r^2

where 4\pi r^2 is the surface of the Gaussian sphere of radius r.

In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than

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E4\pi r^2 = 0\\\rightarrow E = 0

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B)

Here we apply again Gauss Law:

E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}

In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so

r=R=0.18 m

The charge contained within the Gaussian sphere is therefore

q=-29.0 \mu C = -29.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C

And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is

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For this part again, we apply Gauss Law:

E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}

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q=-29.0 \mu C = -29.0\cdot 10^{-6}C

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E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C

And again, this is radially inward, so according to the sign convention asked in the problem,

E=1.2\cdot 10^7 N/C

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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