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sashaice [31]
3 years ago
7

A balloon is negatively charged by rubbing and then clings to a wall. does this mean that the wall is positively charged? why or

why not?
Physics
2 answers:
s2008m [1.1K]3 years ago
5 0

Answer:

When you rub the ballon, it steals electrons from the surface of the thing that is used to rub it.

Now, the electrons are un the surface of the ballon, and when you put it near the wall, the electrons of the wall are repelled by the large amount of electrons in the ballon, leaving only the positive charge in the surface of the wall, this positive surface is atracted by the ballon (and the ballon by the positive surface) and this is why the ballon clongs to the wall.

Oliga [24]3 years ago
3 0
<span>No. The wall is neutral but when you rub the balloon it gains electrons making it negative. When it goes near the wall the electrons in the wall repel so that the protons which are positive attract to the negative balloon. It falls because the wall will take electrons from the balloon.

Hope this helps!</span>
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According to the periodic table, the average atomic mass of helium is A) 2 amu. B) 2.0026 amu. C) 4.0026 amu. D) 6.0026 amu.
RUDIKE [14]

Answer:

C) 4.0026 amu

Explanation:

Helium (He) is the 2nd element on the Periodic Table.

It's neutral atom has 2 protons and 2 electrons.

It is in the 1st period and the 18th row.

It is also a Noble gas.

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3 years ago
How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^
arlik [135]

Answer:

How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^2)?

Explanation:

<em>1700N </em>

<em> Mass multiplied by acceleration gives you the amount of force needed for it.</em>

3 0
3 years ago
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When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the
MaRussiya [10]

Answer:

(a) \alpha=-111.26rad/s

(b) s=4450.6in

(c) 8.66in

Explanation:

First change the units of the velocity, using these equivalents 1rev=2\pi rad and 1 min =60s

4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s

The angular acceleration \alpha the time rate of change of the angular speed \omega according to:

\alpha=\frac{\Delta \omega}{\Delta t}

\Delta  \omega=\omega_i-\omega_f

Where \omega_i is the original velocity, in the case the velocity before starting the deceleration, and \omega_f is the final velocity, equal to zero because it has stopped.

\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s

b) To find the distance traveled in radians use the formula:

\theta = \omega_i t + \frac{1}{2} \alpha t^2

\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad

To change this result to inches, solve the angular displacement \theta for the distance traveled s (r is the radius).

\theta=\frac{s}{r} \\s=\theta r

s=890.12(5)=4450.6in

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

\frac{890.12}{2\pi}=141.6667

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance  between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle \gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120 is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which  is also the net displacement):

c^2=a^2+b^2-2abcos(\gamma)

c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

4 0
4 years ago
How to derive the fourth equation of motion?
Leya [2.2K]

Answer:

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Explanation:

v= u + at

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