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Leto [7]
3 years ago
8

The atoms on the left end of an ice cube are vibrating _____ the atoms on the right end of the same ice cube.

Physics
1 answer:
boyakko [2]3 years ago
5 0
At the same speed as the other ice cube
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The Hubble constant, $H_{0}$, is the ratio of the recessional velocity of a galaxy to its distance. The Hubble constant is estim
Sergeeva-Olga [200]

Answer:

14 billion light years away

Explanation:

v = Recessional velocity = 70km/s/million parsec

D = Proper distance

Hubble constant

70\ km/s=3260000\ ly\\\Rightarrow 1\ km/s=\dfrac{3260000}{70}\ ly

From Hubble's law we have the relation

v=H_0D\\\Rightarrow D=\dfrac{v}{H_0}\\\Rightarrow D=\dfrac{3260000}{70}\times 300000}\\\Rightarrow D=13971428571.42857\ ly

The distance in light years of the galaxy is 14 billion light years away

5 0
3 years ago
A 70.0Kg cyclist Rides a 15.0kg
Inessa [10]

Answer:

1275     kg*m/s

Explanation:

We'll use the momentum equation:

p=mv

where:

p = momentum

m = mass

v = velocity

Since we're doing the magnitude of momentum of the system, we'll add the mass of the cyclist and the mountain bike together:

70.0 kg + 15.0kg=85.0 kg

Given that, we can now substitute our given values into the momentum equation:

p=mv\\p=85.0kg*15.0m/s

Our final answer is:

p= 1275 kg*m/s

4 0
2 years ago
What do nebulae have the most in common with?
alexdok [17]
Most commonly the answer is A, the sun. A nebula is a nursery for stars. The Sun itself was created in a nebula. Also,you do not need college physics to understand this question


3 0
3 years ago
The distance between two charged objects is doubled. What happens to the electrostatic force between the two?a)It will double.b)
zzz [600]

Answer:

d) It will be cut to a fourth of the original force.

Explanation:

The magnitude of the electrostatic force between the charged objects is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the separation between the two objects

In this problem, the initial distance is doubled, so

r' = 2r

Therefore, the new electrostatic force will be

F=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}(k\frac{q_1 q_2}{r^2})=\frac{1}{4}F

So, the force will be cut to 1/4 of the original value.

5 0
3 years ago
Question 15 (3.33 points) Solve: What work is done when 3.0 C is moved through an electric potential difference of 1.5 V?
horsena [70]

Answer:

=_____________________________________

Explanation:

8 0
3 years ago
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