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2 years ago

1.Do parallel universe exist? How?

Physics

1 answer:

Harlamova29_29 [7]2 years ago

7 0

Answer:

Some physicists believe in a flatter version of multiple universes. ... But if the universe began at a finite point, as nearly every physicist agrees that it did, an alternate version of you likely doesn't exist, according to astrophysicist Ethan Siegel's 2015 Medium article.

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Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

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

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

Under certain circumstances, potassium ions (K+) in a cell will move across the cell membrane from the inside to the outside. Th

choli [55]

Answer:

$1.368\times 10^{-20}\ J$

Explanation:

q = Charge in the potassium ion = $19e-18e$

e = Charge of electron = $1.6\times 10^{-19}\ C$

$V_2-V_1$ = Change in potential = $0-(-85.5\times 10^{-3})$

Change in electric potential is given by

$E=q(V_2-V_1)\\\Rightarrow E=(19e-18e)(0-(-85.5\times 10^{-3})\\\Rightarrow E=1.6\times 10^{-19}\times 85.5\times 10^{-3}\\\Rightarrow E=1.368\times 10^{-20}\ J$

The energy is $1.368\times 10^{-20}\ J$

3 0

3 years ago

Plasticity refers to the

olasank [31]

the brain's ability to change throughout the lifespan as a result of experience

5 0

3 years ago

A single sports fan is capable of yelling at an intensity level of 80 dB from a given distance. If 10,000 similar fans were all

riadik2000 [5.3K]

Answer: option "A" is correct

Explanation: the closet to understand the intensity level will be from 80db (decibels) upwards. So option a which is 84db is the answer. Thank you.

4 0

3 years ago

Read 2 more answers

A cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5 m/s2 for 8.0

Katarina [22]

Answer:

V = 15m/s

Explanation:

Given the following data;

Initial velocity = 3m/s

Time = 8secs

Acceleration = 1.5m/s²

To find the final velocity, we would use the first equation of motion;

V = U + at

Substituting into the equation, we have

V = 3 + 1.5*8

V = 3 + 12

V = 15m/s

6 0

2 years ago