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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.
Physics
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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kumpel [21]

Answer:

a = 5.1\ m/s^2

Explanation:

Given,

The angle of the slide=42^\circ

The mass of the child is= m

coefficient of friction = 0.20

when she slides down now apply Newton's law

ma =mg \sin\theta - f

ma = mg\sin \theta -\mu mg \cos \theta

therefore the acceleration

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

a=g[\sin \theta -\mu \cos \theta]

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

a = 5.1\ m/s^2

hence, the magnitude of acceleration during her sliding is equal to  a = 5.1\ m/s^2

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2 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:

2ah=v_f^2-v_0^2

Where:

\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}

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

2ah=v_f^2

The acceleration is the acceleration due to gravity:

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:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

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:

\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2

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:

70.67\frac{m^2}{s^2}=v^2

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}

Therefore, the velocity is 8.4 m/s

7 0
1 year ago
Which of the following is not a process that forms metallic ore
Kamila [148]

 

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

placer deposits

fossil compaction

hydrothermal solutions

igneous processes

 

Actually among all, I have never encountered an ore that formed due to fossil compaction. I suppose we can get minerals such as marble or lime but not ores. So the answer is:

 

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

3 0
3 years ago
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