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3 years ago

What is dark energy?

1 answer:

4 0

Dark energy is a theoretical repulsive energy that causes the acceleration to the expansion of the universe. Among the choices, the nearest answer would be D. Dark energy can also be defined as a new form of energy, that is a dynamic field that fills up space but has an effect opposite to that of normal energy.

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On the position-time graph, what is the average velocity between 5 s and 10 s?

Stells [14]

For this case, we observe that in the interval of 5 to 10 seconds the behavior of the function is linear. Therefore, the average speed in that interval will be given by the slope of the line. This can be calculated as follows
m = (y2-y1) / (x2-x1)
Substituting the values of points A and C
m = ((30) - (10)) / ((10) - (5)) = 4
answer
the average velocity between 5 s and 10 s is 4m / s
B. +4.0 m / s

5 0

3 years ago

Which set of equipment would be most useful to determine the density of a liquid?

NemiM [27]

Triple beam balance/scale to find mass, and graduated cylinder to find volume

3 0

4 years ago

How to do this question?

Zigmanuir [339]

Answer:

First.

Find double diffrenciation of all equation
Then put the value of t in the differenciated equations

3 0

3 years ago

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on

Romashka [77]

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$\frac{1}{4\pi\spsilon_0}=9\times10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35\times10^{-4}$ N

So, from equation (i)

$1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}$

$\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$\frac{q_1+q_2}{2}$

d=20cm=0.2m, and $F_c=1.406\times10^{-4}$ N

So, from equation (i)

$1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}$

$\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}$

$\Rightarrow q_1+q_2=\pm5\times 10^{-8}$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)$