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maks197457 [2]
3 years ago
7

What is the term for the positive charge in an atom?

Physics
2 answers:
kotykmax [81]3 years ago
7 0
The term for the positive charge in an atom is specifically called, "A proton"
dlinn [17]3 years ago
3 0
The answer is A. Proton. Protons have the opposite charge to electrons, which have negative charges. 
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State the Pascal's principle <br>​
krok68 [10]

Answer:

Pascal's principle, also called Pascal's law, in fluid (gas or liquid) mechanics, statement that, in a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container

8 0
3 years ago
An electromagnetic wave has a frequency of 5.0 x 1014 Hz. What is the
andrezito [222]

Answer:

c

Explanation:

wavelength = speed of light/ frequency

= (3x 10^8 m/s)/(5.0 x 10^14 Hz)

= 6.0 x 10^-7 m

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3 years ago
Which has a greater force: a semi-truck at rest or a moving bicycle?
Wittaler [7]

Although the semi truck certainly has a larger mass, it is not in motion and therefore does not have any momentum. The bicycle however has both mass and velocity and therefore has the larger momentum of the pair.

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3 years ago
Read 2 more answers
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
PLEASE PLEASE HELP
pav-90 [236]

if in series one lightbulb burns out the rest are unable to turn on.

In parallel a single light bulb burns out any other light bulbs are able to work.

Parallel is the best to use during holidays.

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3 years ago
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