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svp [43]
2 years ago
14

Que es la expansión del universo?

Physics
1 answer:
anastassius [24]2 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 hammer strikes one end of a thick iron rail of length 5.60 m. A microphone located at the opposite end of the rail detects two
never [62]

Answer:

A) The wave that travels through the rail reaches the microphone first.

B) separation in time between the arrivals of the two pulses is 0.01539 seconds.

Explanation:

Detailed explanation and calculation is shown in the image below

4 0
3 years ago
The Coulomb force between two charges q1 and q2 at separation r in air is F. If half of the separation is filled with medium of
vodomira [7]

Answer:

The new Coulomb force is q₁q₂/9πε₀r²

Explanation

The coulomb force between the two charges q₁ and q₂ at a distance r in air is given by F = q₁q₂/4πε₀r².

Now, let us assume the material of dielectric constant κ = 9 is placed between them on the side of the q₁ charge. The value of its effective charge is now q₃ = q₁/κ at a distance of d = r/2 from the q₂ charge.

Since we have air between q₂ and q₃, the coulomb force between them is

F' = q₂q₃/4πε₀d²

= q₂(q₁/κ)/4πε₀(r/2)²

=  4q₂q₁/κ4πε₀r²

= 4/κ(q₂q₁/4πε₀r²)

= 4/9 × (q₂q₁/4πε₀r²)

= q₁q₂/9πε₀r²

So, the new Coulomb force is q₁q₂/9πε₀r²

3 0
3 years ago
An open organ pipe 30 cm long and a closed organ pipe 23 cm long, both of same diameter , are each sounding its first overtone ,
cupoosta [38]

First overtone of open organ pipe is given as

f_{1o} = \frac{v}{L_1 + 2e}

first overtone of closed organ pipe is given as

f_{1c} = \frac{3v}{4(L_2 + e)}

now they are in unison so we will have

\frac{v}{L_1 + 2e} = \frac{3v}{4(L_2 + e)}

\frac{1}{30 + 2e} = \frac{3}{4(23 + e)}

90 + 6e = 92 + 4e

e = 1 cm

so end correction of both pipes is e = 1 cm

8 0
2 years ago
PLEASE HELP DUE BEFORE 11:30 TODAY!!!!
user100 [1]

Answer:

D is not the a vector quantities

4 0
3 years ago
A shopping cart given an initial velocity of 2.0 m/s undergoes a constant acceleration to a velocity of 13 m/s. What is the magn
olga55 [171]

Answer:

The acceleration is a = 2.75 [m/s^2]

Explanation:

In order to solve this problem we must use kinematics equations.

v_{f} = v_{i} + a*t\\

where:

Vf = final velocity = 13 [m/s]

Vi = initial velocity = 2 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Now replacing:

13 = 2 + (4*a)

(13 - 2) = 4*a

a = 2.75 [m/s^2]

5 0
3 years ago
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