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baherus [9]
3 years ago
7

A particle with a charge of -4.0 μC and a mass of 3.2 x 10-6 kg is released from rest at point A and accelerates toward point B,

arriving there with a speed of 72 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VB is less than VA, then give the answer as a negative number.
Physics
1 answer:
Assoli18 [71]3 years ago
4 0

Answer:

V_B-V_A=-20736-0=-20736volt

Explanation:

We have given charge on the particle q=-4\mu C=-4\times 10^{-6}C

Mass of the charge particle m=3.2\times 10^{-6}kg

From energy of conservation kinetic energy will be equal to potential energy

So at point A

\frac{1}{2}mv^2=qV

At point a velocity is zero

So \frac{1}{2}(3.2\times10^{-6} )0^2=-4\times 10^{-6}V_a

V_A=0volt

At point B velocity will be 72 m/sec

So \frac{1}{2}\times 3.2\times 10^{-6}72^2=-4\times 10^{-6}V_b

V_B=-20736volt

So V_B-V_A=-20736-0=-20736volt

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Answer:

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A systematic error occurs as a result of the instrument used in carrying out and experiment. These errors are a result of small fluctuations in the measurement properties of the instrument. This happens when the instrument departs from non-ideal situations, for example as a result of physical expansion or change in temperature. For instance, let the resistance be measured to be up to 10 Ω ± 1 Ω

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The difference between Lincoln's plan of reconstruction and the Radical Republicans' is that Lincoln wanted to reunite the union
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Answer:

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Which biome has the most variable year round temperature
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You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 77.2 kg hop on board for a ride
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To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.

By Hook's law we know that force is defined as,

F= kx

Where,

k = spring constant

x = Displacement change

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k= \frac{F}{x}

k = \frac{mg}{x}

k= \frac{77.2*9.8}{0.0637}

k = 11876.92N/m

Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m

PART B) In the case of speed we can obtain it through the period, which is given by

T = \frac{2\pi}{\omega}

Re-arrange to find \omega,

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{2.14}

\omega = 2.93rad/s

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t_T = 10*T

t_T = 10 *2.14s

t_T = 21.4s

Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s

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