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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:

You might be interested in
An 8 kg mass moving at 8 m/s collides with a 6 kg mass
steposvetlana [31]

Answer:

10 m/s

Explanation:

Momentum before collision = momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(8 kg)(8 m/s) + (6 kg)(6 m/s) = (8 kg)(5 m/s) + (6 kg) v

64 kg m/s + 36 kg m/s = 40 kg m/s + (6 kg) v

60 kg m/s = (6 kg) v

v = 10 m/s

3 0
3 years ago
Why is this event important to include in the biography?
nadya68 [22]

Answer:

D: It shows that Frida Kahlo used art to cope with her pain.

Explanation:

Within the text given it shows her emotions being lonely, immobile and in pain. But it all shows her asking her father for art which states that art is her sort of relief and happy place.

8 0
2 years ago
Read 2 more answers
When gases, liquids, or solids are in contact with a moving object, the flow of
Gwar [14]

Answer:

Heat

Friction is what causes heat.

Brainliest always helps.

5 0
2 years ago
(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 48.0 µC. What is the voltage of the
frosja888 [35]

Answer:

a.6 V

b.24 V

Explanation:

We are given that

a.C=8\mu F=8\times 10^{-6} F

1\mu =10^{-6}

Q=48\mu C=48\times 10^{-6} C

We know that

V=\frac{Q}{C}

Using the formula

V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V

b.Q=192\mu C=192\times 10^{-6} C

V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V

8 0
2 years ago
A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget
mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, m_{1} = 4 kg

Speed, v_{1} = 5 m/s

m_{2} = 1 kg

v_{2} = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2

By plugging in the values we get,

KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)

KE = 40J + KE(lost)

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0
3 years ago
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