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mojhsa [17]
4 years ago
14

How many meters are there in 4.80 ly ?

Physics
1 answer:
Masja [62]4 years ago
6 0
Since each light year is approximately 9 trillion kilometres, 4.80 light years is 43.2 trillion kilometres, or 43,200,000,000,000,000 metres
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ID: A
postnew [5]

Answer: Velocity=8.26m/s

Explanation:

Acceleration=Finial velocity (V) - Initial velocity (u) ÷ Time

that is, a=v-u/t

a=1.2m/s², v=?, u=5.5m/s, t=2.3s

From a=v-u/t, make v the subject of the formula

v=at + u

v=(1.2* 2.3) + 5.5

v=2.76+5.5

v=8.26m/s

4 0
3 years ago
Which of the following are true? Select all that apply. The net electric field at any location inside a block of copper is zero
Agata [3.3K]

Answer:

1) The net electric field at any location inside a block of copper is zero if the copper block is in equilibrium.

2) In equilibrium, there is no net flow of mobile charged particles inside a conductor.

3) If the net electric field at a particular location inside a piece of metal is not zero, the metal is not in equilibrium.

Explanation:

1) and 3) A block of copper is a conductor. The charged particles on a conductor in equilibrium are at rest, so the intensity of the electric field at all interior points of the conductor is zero, otherwise, the charges would move resulting in an electric current.

2) The charged particles on a conductor in equilibrium are at rest.

6 0
3 years ago
To identify whether the forces are balanced or not, you need to whet
jok3333 [9.3K]

Answer:

what are you asking?

Explanation:

7 0
3 years ago
Elements are arranged in the periodic table based on various patterns. For example, the element magnesium (Mg) A. has a higher a
Sati [7]

The right answer is A just did the question.


7 0
3 years ago
In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs
bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

V=\frac{I}{nAq}

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

\frac{I}{q} = 1.2*10^{18}

A= 1.3*10^{-8}m^2

n=6.3*10^{28} e/m^3

\omicron{O} = 1.2*10^{-4}(m/s)(N/c) Mobility

We can find the drift velocity replacing,

V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}

V= 1.465*10^-3m/s

The electric field is given by,

E= \frac{V}{\omicron{O}}

E=\frac{1.465*10^-3}{1.2*10^{-4}}

E=12.2V/m

7 0
4 years ago
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