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svp [43]
3 years ago
14

Que es la expansión del universo?

Physics
1 answer:
anastassius [24]3 years ago
3 0

Answer:

La expansión no es más que el incremento con el tiempo de la distancia entre cualquier par de galaxias lejanas. Se suele utilizar para representar este hecho la analogía de un globo donde hemos pintado una serie de puntos a modo de galaxias.

Explanation:

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A point charge of 6.0 nC is placed at the center of a hollow spherical conductor (inner radius = 1.0 cm, outer radius = 2.0 cm)
egoroff_w [7]

Explanation:

The given data is as follows.

             q = 6.0 nC = 6 \times 10^{-9} C

         inner radius (r) = 1.0 cm = 0.01 m   (as 1 cm = 100 m)

So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.

Formula to calculate the charge density is as follows.

            \sigma = \frac{q_{in}}{A} .......... (1)

Since, area of the sphere is as follows.

               A = 4 \pi r^{2} ........... (2)

Hence, substituting equation (2) in equation (1) as follows.

      \sigma = \frac{q_{in}}{4 \pi r^{2}}

                   = \frac{6 \times 10^{-9} C}{4 \times 3.14 \times (0.01)^{2}}            

                   = 0.477 \times 10^{-5}

or,               = 4.77 \mu C/m^{2}

Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 \mu C/m^{2}.

5 0
3 years ago
Roger drives his car at a constant speed of 80 km/hr. How far can he travel in 2 hrs. and 30 minutes?
bonufazy [111]

Answer:

200 km/hr

Explanation:

Since he goes 80km per hour, multiply this by 2.5 or two and a half hours.

80 x 2.5 = 200 km/hr.

6 0
3 years ago
The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross-sectional area of the solenoi
Novay_Z [31]

Answer: 0.43 V

Explanation:

L = [μ(0) * N² * A] / l

Where

L = Inductance of the solenoid

N = the number of turns in the solenoid

A = cross sectional area of the solenoid

l = length of the solenoid

7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24

1.752*10^-3 = 4π*10^-7 * 202500 * A

1.752*10^-3 = 0.255 * A

A = 1.752*10^-3 / 0.255

A = 0.00687 m²

A = 6.87*10^-3 m²

emf = -N(ΔΦ/Δt).........1

L = N(ΔΦ/ΔI) so that,

N*ΔΦ = ΔI*L

Substituting this in eqn 1, we have

emf = - ΔI*L / Δt

emf = - [(0 - 3.2) * 7.3*10^-3] / 55*10^-3

emf = 0.0234 / 0.055

emf = 0.43 V

6 0
3 years ago
CAN someone help ASAP?
lyudmila [28]

Answer:someone help me

Explanation:

7 0
3 years ago
A student claims "everything falls at the same acceleration rate on the Moon, where there is no air or friction," how would you
Alexeev081 [22]

Now let’s say you’re on the Moon. If you were to drop a hammer and a feather from the same height, which would hit the ground first?

Trick Question! On the moon both objects would hit the ground at the same time. On Earth, the hammer lands first.

So yeah, the student is right. Galileo gave us this theory long ago.

5 0
3 years ago
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