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slavikrds [6]
3 years ago
9

In a parallel circuit the current A stays in one path B splits and goes through two components C makes one circle

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
8 0

Answer:

B splits and goes through two components

Explanation:

- A series circuit is a circuit in which the components are all connected along the same branch: as a result, the current flowing through the components is the same, while the sum of the potential differences across each component is equal to the emf of the battery

- A parallel circuit is a circuit consisting of separate branches, so that each branch has a potential difference equal to the emf of the battery. As a result, in such a circuit the current in the circuit splits and goes through the different branches/components.

So, the correct answer is

B splits and goes through two components

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I NEED HELP PLEASE, THANKS! :)
makkiz [27]

Answer:

They move outwards.

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Explanation:

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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
leonid [27]

Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

(b): F_g = 3.6125\times 10^{-47}\ \rm N.

(c): \dfrac{F_e}{F_g}=2.27\times 10^{39}.

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges q_1 and q_2 respectively is given by

F_e = \dfrac{k|q_1||q_2|}{r^2}

where,

  • k = Coulomb's constant = 9\times 10^9\ \rm Nm^2/C^2.
  • r = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, q_1 = +1.6\times 10^{-19}\ C.

The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

These two are separated by the distance, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

where,

  • G = Universal Gravitational constant = 6.67\times 10^{-11}\ \rm Nm^2/kg^2.
  • r = distance of separation between the masses.

For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

6 0
3 years ago
Mathphys Help help help
Keith_Richards [23]

Answer:

1030 mph

Explanation:

The new velocity equals the initial velocity plus the wind velocity.

First, in the x (east) direction:

vₓ = 335 mph + 711 cos 19° mph

vₓ = 1007 mph

And in the y (north) direction:

vᵧ = 0 mph + 711 sin 19° mph

vᵧ = 231 mph

The net speed can be found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1007 mph)² + (231 mph)²

v ≈ 1030 mph

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2. A bus drove 8 meters East, then turned to drive 8 meters North, then 2 meters
Sphinxa [80]

Answer:

Long question good luck:)

Explanation:

3 0
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