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Sergio [31]
3 years ago
9

In Figure 18.8 the grounding wire is removed first, followed by the rod, and the sphere is left with a positive charge. If the r

od were removed first, followed by the grounding wire, would the sphere be left with a charge? Account for your answer.

Physics
1 answer:
Elena-2011 [213]3 years ago
3 0
<span>In Figure 18.9, the ground wire is first removed, then the rod. This leaves a positive charge on the sphere. If you remove the rod first and then remove the ground wire, there will be no charge left on the sphere. When the rod is removed, the repulsive force caused by the presence of the rubber rod no longer exists. As the wire is still attached, free electrons enter the sphere from the ground until the sphere is neutral again.</span>
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Answer:

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4 years ago
Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c
Mariulka [41]

Answer:

A. v_{3}=12.17m/s

B. v_{car}=6.3m/s\\v_{truck}=-6.3m/s

C. ΔK=-4.13x10^3J

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg

A. Since the two vehicles become entangled the final mass is:

m_{3}=102kg+103kg=205kg

From linear momentum we got that:

p_{1}=p_{2}

m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}

v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}

v_{3}=12.17m/s

B. The change in velocity of both vehicles are:

For the car

v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s

For the truck

v_{truck}=12.17m/s-18.5m/s=-6.3m/s

C. The change in kinetic energy is:

ΔK=K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})

ΔK=\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J

ΔK=-4.13x10^{3}J

6 0
4 years ago
The question is on the picture.
Jet001 [13]
Between noon and 2 pm, the amount of water in the rain gauge decreased.
This can be caused by evaporation, which turns water into water vapor.
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So the answer is A. evaporation.
5 0
3 years ago
Read 2 more answers
Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a
nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

Q = \int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi

Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {r^3} \, dr * d\theta* d\phi

Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi

Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \,  d\phi

Q = \frac{\pi^2 r^4}{2}}

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

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3 years ago
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Answer:

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3 years ago
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