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Juliette [100K]

3 years ago

8

Two balloons are charged with an identical quantity and type of charge: -4 nc They are held apart at a separation distance of 70

cm. Determine the magnitude of the electrical force of repulsion between them.

1 answer:

blagie [28]3 years ago

8 0

Answer:

29.4 uN

Explanation:

The electric force between two charges can be calculated using Coulomb's Law. According to this law the force between two point charges is given as:

$F=k\frac{q_{1} q_{2} }{r^{2}}$

where k is a proportionality constant known as the Coulomb's law constant. Its value is $9 \times 10^{9}$ Nm²/C²

r = distance between charges = 70 cm = 0.7 m

q1 = q2 = 4nC = $4 \times 10^{-9}$ C

The negative sign indicates that the charges are negative. In the formula we will only use the magnitude of the charges.

Using these values in the formula, we get:

$F=9 \times 10^{9} \times \frac{4 \times 10^{-9} \times 4 \times 10^{-9}}{0.7^{2}}\\\\ F=2.94\times10^{-7} N\\\\ F=29.4 \times 10^{-6} N\\\\ F=29.4 \mu N$

Therefore, the magnitude of repulsive force between the given charges will be 29.4 uN

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Answer:

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Explanation:

Given,

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therefore the acceleration

$a=g \sin\theta -\mu gcos\theta$

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$a=9.8\times [\sin 42^\circ -0.2\times \cos 42^\circ]$

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3 years ago

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6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

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Where:

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If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

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$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

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$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

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Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

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Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

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Kamila [148]

I believe that this question has the following choices to choose from:

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<span>fossil compaction (answer)</span>

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3 years ago

Read 2 more answers