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3 years ago

a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?

Physics

1 answer:

Verdich [7]3 years ago

6 0

The distance between the two charges is $0.3 m$

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

$q_1 = 8\cdot 10^{-5} C$

$q_2 = 3\cdot 10^{-5} C$

$F=2.4\cdot 10^2 N$

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

$r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m$

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A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first

Dominik [7]

Answer:

a) $V=14.01 m/s$

b) $V=8.86 \, m/s$

c) $t = 2.33s$

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

$E_m = E_k +E_p$

That is, mechanical energy is equal to the sum of potential and kinetic energy, and the value of this $E_m$ mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

$E_k = \frac{1}{2} m \, V^2$

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

$E_p= m\, g\, h$

where g is the acceleration due to gravity ( $g=9.81 \, m/s^2$ ) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which $h=0$ , a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

$E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m$

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set $h=0$

$E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2$

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

$E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\$

And so we have found the velocity of the ball as it hits the floor.

$V = \sqrt[]{2g\cdot 10m}=14.01\, m/s$

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

$V = \sqrt[]{2g\cdot h}$

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

$V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s$

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average? It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

$V_{avg}\cdot t=h\\t=h/V_{avg}$

what's the average speed when the ball is descending?

$V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s$

so the time it takes the ball to go down is:

$t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\$

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

$V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s$

and the time it takes to go up is: $t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s$

When we add both times , we get:

$t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s$

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Lithium-Ion batteries are commonly used in portable electronics and electric vehicles. These rechargeable batteries have two electrodes: one that's positively charged and contains lithium and another negative one that's typically made of graphite.

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If the pressure of a substance is increased during a boiling process, will the temperature also increase, or will it remain cons

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Answer:

on increasing pressure, temperature will also increase.

Explanation:

Considering the ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of the gas.

P ∝ T

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Also, using Gay-Lussac's law,

$\frac {P_1}{T_1}=\frac {P_2}{T_2}$

Thus, on increasing pressure, temperature will also increase.

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17. An uncharged metal rod brought close to but not touching the cap of a charged electroscope caused decrease in the divergence

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