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3 years ago

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The work done by an external force to move a -8.50 μC charge from point a to point b is 6.10×10−4 J . If the charge was started

from rest and had 1.50×10−4 J of kinetic energy when it reached point b, what must be the potential difference between a and b?

1 answer:

bekas [8.4K]3 years ago

5 0

Answer:

-54.12 V

Explanation:

The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

$W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J$

The change in electrostatic potential energy $\Delta U$ , of one point charge q is defined as the product of the charge and the potential difference.

$\Delta U=qV\\V=\frac{\Delta U}{q}\\V=\frac{4.60*10^{-4}J}{-8.50*10^{-6}C}\\V=-54.12 V$

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Answer:

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Wind energy is a form of kinetic energy in motion.

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Explanation:

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3 years ago

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Answer:

F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

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V_{water} = 1.09 10⁻³ m³

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ρ = m / V

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m = 1000 1.09 10⁻³

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P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2 ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

P₂ = P_{atm}

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let's use kinematics to find the speed of a drop when it reaches the container y = 0

v² = v₀² - 2 g (y-y₀)

v = $\sqrt{0 -2 g ( 0-y_o)}$

let's calculate

v = √(2 9.8 40.5)

v = 28.17 m / s

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we substitute in Bernoulli's equation

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MakcuM [25]

Answer:

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