Answer:
The age of the universe would be 9.9 billion years
Explanation:
We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.
Thus, the time it has taken for the galaxies to reach their current separations is:
and from Hubble's Law:
Therefore:
With the given value for the Hubble's constant we have:
and thus,
Answer:
it loses engry it follows difernt paths
Explanation:
The original width was 94.71 cm
<span>The area decreased 33.1% </span>
<span>The equation for the final size is </span>
<span>2X^2 = 1.2 m^2 </span>
<span>X^2 - 0.6 m^2 </span>
<span>X^2 = 10000 * .6 cm </span>
<span>X = 77.46 cm (this is the width) </span>
<span>The length is 2 * 77.46 = 154.92 cm </span>
<span>The original length was 154.92 + 34.5 = 189.42 cm </span>
<span>The original width was 189.42 / 2 = 94.71 cm </span>
<span>The original area was 94.71 * 189.92 = 17939.9 cm^2 </span>
<span>The new area is 79.46 * 154.92 = 12000.1 cm^2 </span>
<span>The difference between the original and current area is 17939.9 - 12000.1 = 5939.86 cm^2 </span>
<span>The percentage the area decreased is 5939.86 ' 17939.9 = 33.1%</span>
Answer:
0.25 m.
Explanation:
We'll begin by calculating the spring constant of the spring.
From the diagram, we shall used any of the weight with the corresponding extention to determine the spring constant. This is illustrated below:
Force (F) = 0.1 N
Extention (e) = 0.125 m
Spring constant (K) =?
F = Ke
0.1 = K x 0.125
Divide both side by 0.125
K = 0.1/0.125
K = 0.8 N/m
Therefore, the force constant, K of spring is 0.8 N/m
Now, we can obtain the number in gap 1 in the diagram above as follow:
Force (F) = 0.2 N
Spring constant (K) = 0.8 N/m
Extention (e) =..?
F = Ke
0.2 = 0.8 x e
Divide both side by 0.8
e = 0.2/0.8
e = 0.25 m
Therefore, the number that will complete gap 1is 0.25 m.