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3 years ago

Which of the following is evidence that the universe began with the big bang theory

1 answer:

6 0

<span>There are several main pieces of evidence that support the Big Bang theory. One is the fact that the universe is expanding, proven with something called red shift. The second is something called cosmic microwave background radiation. The third is the abundance of different elements in the universe.

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Find the average acceleration of a northbound subway train that slows down from 12 m/s to 9.6 m/s in 0.8 seconds.

iren2701 [21]

Answer:

$a=-3\ m/s^2$

Explanation:

Given that,

Initial speed of a train, u = 12 m/s

Final speed of a train, v = 9.6 m/s

Time taken, t = 0.8 s

We need to find the average acceleration of the train. We know that the rate of change of velocity is equal to acceleration. So,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{9.6-12}{0.8}\\\\a=-3\ m/s^2$

So, the average acceleration of the train is equal to $-3\ m/s^2$ .

8 0

3 years ago

Plz help i'm desperate! Why are all people on Earth members of the same biological Species?

Alex Ar [27]

Answer:

D.) All people can mate with one another and have children

Explanation:

7 0

3 years ago

A + 8.42 nC point charge and a - 3.75 nC point charge are 2.73 cm apart. What is the electric field strength at the midpoint bet

aev [14]

Given:

The charge q1 = 8.42 nC

The charge q2 = -3.75 nC

The distance between the charges is d = 2.73 cm

To find the electric field strength at the midpoint between the two charges.

Explanation:

The distance between the charges and the midpoint is d' =d/2

The electric field strength can be calculated by the formula

$E\text{ = }\frac{k|q|}{d^{\prime2}}$

The electric field strength at the midpoint due to the charge q1 will be

$\begin{gathered} E_1=\frac{9\times10^9\times8.42\times10^{-9}\text{ C}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =\text{ 4.07}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint due to the charge q2 will be

$\begin{gathered} E_2=\frac{9\times10^9\times3.75\times10^{-9}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =1.8\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the charges will be

$\begin{gathered} E=E_1+E_2 \\ =4.07\times10^5+1.8\times10^5 \\ =\text{ 5.87}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the two charges is 5.87 x 10^(5) m.

4 0

1 year ago

On the position vs time graph a curved line represents ____________.

Art [367]

The velocity is changing.

6 0

3 years ago

A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with

NNADVOKAT [17]

Answer: 1.59 sec

Explanation:

The kinetic energy contained in the rotation of the cylinder is 1/2 m v^2

The kinetic energy of translation is also 1/2 m v^2

so the total energy is m v^2

The force applied is mg sin (theta)

= m x 9.8 x 1/2

= 4.9 m

Now equate

F x d = m v^2

4.9 m x 3.1 = m v^2

v^2 = 4.9 x 3.1

v = sqrt(4.9 x 3.1) = 3.9 m/s

Acceleration.

V^2 = 2 a s

4.9 x 3.1 = 2 x a x 3.1

4.9 = 2 a

a = 2.45 m/s^2

Time

T = v/a

= 3.9/2.45

= 1.59 sec

6 0

4 years ago