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Sedaia [141]
3 years ago
8

What happens when a negatively charged object A is brought near a neutral object B?

Physics
2 answers:
Andrews [41]3 years ago
6 0
Object B gets a positive charge
Eva8 [605]3 years ago
5 0
<span>Object B stays neutral but becomes polarized.

Hope this helps.

~Jurgen</span>
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In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0800 s, during which
Elena-2011 [213]
The initial velocity of the ball is 0. Applying:
v = u + at
v = 0 + 229 x 0.08
v = 18.3 m/s

a)
Vx = Vcos(∅)
Vx = 18.3cos(52.3)
Vx = 11.2 m/s

b)
Vy = Vsin(∅)
Vy = 18.3sin(52.3)
Vy = 14.5 m/s
5 0
4 years ago
The cosmic microwave background essentially looks the same in all directions. This is an example of.
pentagon [3]

In general, all views of the cosmic microwave background are identical. Isotropy is demonstrated by this.

<h3>What exactly is isotropy, for instance?</h3>

The Greek words isos (equal) and tropos, from which the term "isotropy" is derived, mean "uniform in all directions" (way). The material properties of anisotropic materials, such as graphite, differ depending on the direction, in contrast to isotropic materials like glass, which show the same properties in all directions.

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3 0
1 year ago
A negatively charged balloon has 4 μC of charge. How many excess electrons are on this bal- loon? The elemental charge is 1.6 ×
bagirrra123 [75]
Data:

The charge of a body depends on the amount of electrons it gains or loses. Q = n * e, where "Q" is charge, "n" is the number of plus or minus electrons, and "e" is the fundamental charge of an electron 1,6 * 10 ^{-19}C<span>. To know if the body has gained or lost, we look at the signal of its charge, remembering that the electron is negative. The charge of the body is 4 μC (positive), so there is a lack of electrons! 

Q = 4 </span>μC → Q = 4*10^{-6}
e = 1,6 * 10 ^{-19}C
n = ?<span>

We have:
</span>Q = n*e
n =  \frac{Q}{e}
n =  \frac{4*10^{-6}}{1,6 * 10 ^{-19}}
n = 2,5*10^{-6-(-19)}
n = 2,5*10^{-6+19}
\boxed{n = 2,5*10^{13}electrons}
7 0
3 years ago
Please help me I have to send for my teacher​
expeople1 [14]

Answer:

for the fill in the blanks

1- static

2-kinetic

3-coeffiecient

4-opposite to

5-sin theta

6-cos theta

im not sure however what to do with the top part or if its even part of what you need help with

8 0
2 years ago
A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee
Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

s = ut +\frac{1}{2}at^{2}

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

s = \frac{1}{2}gt'^{2}

s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}                     (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

s = v\times t''

s = 343\times t''                                              (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t'                                                                        (3)

Using eqn (2) and (3):

s = 343(3.5 - t')                                                                 (4)

from eqn (1) and (4):

4.9t'^{2} = 343(3.5 - t')

4.9t'^{2} + 343t' - 1200.5 = 0

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

3 0
3 years ago
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