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3 years ago

The Big Bang Theory was supported by the discovery of the

2 answers:

7 0

Two major scientific discoveries provide strong support for the Big Bang theory: • Hubble's discovery in the 1920s of a relationship between a galaxy's distance from Earth and its speed; and • the discovery in the 1960s of cosmic microwave background radiation. Please give brainliest

Alex787 [66]3 years ago

6 0

Answer:

For anyone asking, the answer is (Red shift in electromagnetic spectrum)

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Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =

Semenov [28]

For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
$r_1+r_2=L km$
The second equation is going to the total force acting on the charge q3:
$F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}$
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
$0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}$
Let's solve this for r_1^2:
$0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0$
Now we have a quadratic equation with following parameter:
$a=2\\ b=2L\\ c=-L^2$
We know that two solutions are:
$r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\$
We need a positive solution. When we plug in all the numbers we get:
$r_1=0.915\cdot 10^6$km$

6 0

4 years ago

Two electrons travel towards each other at 0.2 c parallel to the laboratory x-axis. What is the relative velocity of one electro

Leya [2.2K]

1) In the reference frame of one electron: 0.38c

To find the relative velocity of one electron with respect to the other, we must use the following formula:

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

where

u is the velocity of one electron

v is the velocity of the second electron

c is the speed of light

In this problem:

u = 0.2c

v = -0.2c (since the second electron is moving towards the first one, so in the opposite direction)

Substituting, we find:

$u'=\frac{0.2c+0.2c}{1+\frac{(0.2c)(0.2c)}{c^2}}=\frac{0.4c}{1+0.04}=0.38c$

2) In the reference frame of the laboratory: -0.2c and +0.2c

In this case, there is no calculation to be done. In fact, we are already given the speed of the two electrons; we are also told that they travel in opposite direction, so their velocities are

+0.2c

-0.2c

5 0

3 years ago

A competitive go-cart driver is traveling at a speed of 32m/s. He sees a caution flag go up and slows down at a rate of -1.5 m/s

djyliett [7]

Answer:

His final velocity is 15.8 m/s.

Step-by-step explanation:

Given:

Initial velocity of the driver is, $u=32$ m/s

Acceleration of the driver is, $a=-1.5$ m/s²

Time taken to reach final velocity is, $t=10.8$ s.

The final velocity is given using the Newton's equations of motion as:

$v=u+at$ , where, $v$ is the final velocity.

Now, plug in the given values and solve for $v$ .

$v=32-1.5(10.8)\\v=32-16.2=15.8\textrm{ m/s}$

Therefore, his final velocity is 15.8 m/s.

5 0

3 years ago

A storm is moving at 15 km/hr what to do to determine its velocity

WINSTONCH [101]

Check the current weather map for 2 different times, and see where the center of the storm is. That tells you what direction it's moving. With its speed and direction, you have its velocity.

6 0

3 years ago

A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a

iren [92.7K]

Answer:

√(6ax)

Explanation:

Hi!

The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:

x=(1/2)at^2

Then the we need to find the time t' for which the displacement is 3x

3x=(1/2)a(t')^2

Solving for t':

t'=√(6x/a)

Now, the velocity of the motorcycle as a function of time is:

v(t)=a*t

Evaluating at t=t'

v(t')=a*√(6x/a)=√(6*x*a)

Which is the final velocity

Have a nice day!

3 0

3 years ago