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s2008m [1.1K]
2 years ago
5

The question is in the picture

Physics
1 answer:
OLEGan [10]2 years ago
4 0

Answer:

Explanation:

ME initial = ME final

PE initial + KE initial = PE final + KE final

There's no kinetic energy the moment Jorge is at the top of the ramp therefore KE initial = 0

PE initial = PE final + KE final

Assuming the bottom of the ramp is h = 0 we can then say PE final = 0

PE initial = KE final

Write the equations out.

mgh = (.5)mv^2

They tell us potential energy is 6000J therefore that portion of the equation is already "solved"

6000 = (.5)mv^2

Plugin the values

6000 = (.5) 70 v^2

Solve for v

6000 = 35v^2

171.43 = v^2

13.1 = v at the bottom of the slope

Solving for KE we get

(.5) m v^2

Plugin the value from the previous problem and solvee!

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Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on
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Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

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where,

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Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

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q₁q₂ = 4.17 x 10⁻¹²

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The sum of charges is given as:

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using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C   (OR)   q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC   (OR)   q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

<u>B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C</u>

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