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Nat2105 [25]
3 years ago
5

Can you tell from your experiment so far whether the tapes carry a positive charge or a negative charge? Briefly explain your an

swer.
Physics
1 answer:
Softa [21]3 years ago
8 0

Answer:

Explanation:

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charge: positive and negative (commonly carried by protons and electrons respectively). Like charges repel each other and unlike charges attract each other. An object with an absence of net charge is referred to as neutral. Early knowledge of how charged substances interact is now called classical electrodynamics, and is still accurate for problems that do not require consideration of quantum effects.

Electric charge is a conserved property; the net charge of an isolated system, the amount of positive charge minus the amount of negative charge, cannot change. Electric charge is carried by subatomic particles. In ordinary matter, negative charge is carried by electrons, and positive charge is carried by the protons in the nuclei of atoms. If there are more electrons than protons in a piece of matter, it will have a negative charge, if there are fewer it will have a positive charge, and if there are equal numbers it will be neutral. Charge is quantized; it comes in integer multiples of individual small units called the elementary charge, e, about 1.602×10−19 coulombs,[1] which is the smallest charge which can exist freely (particles called quarks have smaller charges, multiples of

e, but they are only found in combination, and always combine to form particles with integer charge). The proton has a charge of +e, and the electron has a charge of −e.

An electric charge has an electric field, and if the charge is moving it also generates a magnetic field. The combination of the electric and magnetic field is called the electromagnetic field, and its interaction with charges is the source of the electromagnetic force, which is one of the four fundamental forces in physics. The study of photon-mediated interactions among charged particles is called quantum electrodynamics.

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seropon [69]
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7 0
2 years ago
Calculate the force required to accelerate<br> a 600 g ball from rest to 14 m/s in 0.1 s.
Evgen [1.6K]

<u>Statement</u><u>:</u>

A force is required to accelerate a 600 g ball from rest to 14 m/s in 0.1 s.

<u>To </u><u>find </u><u>out</u><u>:</u>

The force required to accelerate the ball.

<u>Solution</u><u>:</u>

  • Mass of the ball (m) = 600 g = 0.6 Kg
  • Initial velocity (u) = 0 m/s [it was at rest]
  • Final velocity (v) = 14 m/s
  • Time (t) = 0.1 s

  • Let the acceleration be a.
  • We know the equation of motion,
  • v = u + at

  • Therefore, putting the values in the above formula, we get
  • 14 m/s = 0 m/s + a × 0.1 s
  • or, 14 m/s ÷ 0.1 s = a
  • or, a = 140 m/s²

  • Let the force be F.
  • We know, the formula : F = ma

  • Putting the values in the above formula, we get
  • F = 0.6 Kg × 140 m/s²
  • or, F = 84 N

<u>Answer</u><u>:</u>

The force required to accelerate the ball is 84 N and this force acts along the direction of motion.

Hope you could understand.

If you have any query, feel free to ask.

5 0
3 years ago
A +5.0 uC point charge is placed at the 0 cm mark of a meter stick and a -4.0 °C charge is
viva [34]

Answer:

-1.44*10^11v/m

Explanation:

6 0
3 years ago
meg goes swimming on a hot afternoon. when she comes out of the pool, her foot senses that the pavement is unbearably hot. suppo
Korolek [52]
The next step to take is to used the scientific method to explore the observations she has made in order to discover the facts behind her observations. She should ask questions about the hotness of the pavement and form an hypothesis to support her observation. She should then set up an appropriate experiment to confirm her hypothesis. After the conduction of the experiment, she should analyze her results and then draw a conclusion.<span />
5 0
4 years ago
Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

                     = 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

5 0
3 years ago
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