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3 years ago

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A tiny particle with charge + 5.0 μC is initially moving at 55 m/s. It is then accelerated through a potential difference of 500

V. How much kinetic energy does this particle gain during the period of acceleration?

1 answer:

Vadim26 [7]3 years ago

8 0

Answer:

ΔK.E = 2.5 × 10⁻³ J

Explanation:

Given data in the question, we have:

Charge of the particle, q = 5.0 μC = 5 × 10 ⁻⁶ C

Initial speed of the particle, v = 55 m/s

The potential difference, ΔV = 500 V

Now, the gain in kinetic energy is given as

ΔK.E = q × ΔV

on substituting the values in the above formula, we get

ΔK.E = 5 × 10 ⁻⁶ C × 500 V

or

ΔK.E = 2.5 × 10⁻³ J

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What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?

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Thus, the repulsive force between two pith balls is $8.8\times10^{-4}N$ .

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A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o

Nataly_w [17]

Explanation:

The given data is as follows.

$C_{1} = 1.50 \times 10^{-6} F$

$C_{1} = 3.50 \times 10^{-6} F$

Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}$

= $0.945 \times 10^{-6} F$

C = $1.06 \times 10^{-6} F$

Now, we will calculate the charge across each capacitance as follows.

Q = CV

= $1.06 \times 10^{-6} F \times 2.50 V$

= $2.65 \times 10^{-6} C$

= $2.65 \mu C$

Thus, we can conclude that $2.65 \mu C$ is the charge stored on each given capacitor.

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3 years ago

A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your

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Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

For this we are going to use the equation of distance for an uniformly accelerated movement, that is:

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

Where:

x = Final distance

xo = Initial point

Vo = Initial velocity

a = Acceleration

t = time

We have the following values:

x = 1.66m

xo = 0m (the kangaroo starts from the floor)

Vo = 0 m/s (each jump starts from the floor and from a resting position)

a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)

t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

$1.66 = 0 + 0t + \frac{1}{2}9.8t^2$

$1.66 = 4.9t^2$

$\frac{1.66}{4.9} = t^2$

$\sqrt{0.339} = t\\ t = 0.582s$

It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

t = 1.164s

The kangaroo was 1.164s in the air before returning to Earth

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