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3 years ago

Alternatives to the Big Bang Theory.

1 answer:

7 0

Nearly 14 billion years ago, there was nothing and nowhere. Then, due to a random fluctuation in a completely empty void, a universe exploded into existence. Something the size of a subatomic particle inflated to unimaginably huge size in a fraction of a second, driven apart by negative-pressure vacuum energy. Scientists call this theory for the origin of the universe the Big Bang.

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What is the value of acceleration that the car experiences

Anni [7]

Can u show the problem ?
Acceleration= the rate and change of an object also cars go fast so I think it would be positive acceleration
Hope I helped

7 0

4 years ago

6. An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of t

zubka84 [21]

Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

$d=v\times t$

$d=9.1\times t$ ....(I)

The distance of longitudinal wave

$d=5.7\times (t+71)$ ....(II)

From the first equation

$t=\dfrac{d}{9.1}$

Put the value of t in equation (II)

$d =5.7\times(\dfrac{d}{9.1}+71)$

$\dfrac{9.1d-5.7d}{9.1}=71\times5.7$

$d0.3736=404.7$

$d =1083.24\ km$

Hence, The distance away the center of the earthquake is 1083.24 km.

6 0

3 years ago

The initial and final velocities of two blocks experiencing constant acceleration are respectively −7.45 m/s and 14.9 m/s. (a) T

ikadub [295]

Answer:

a) $a_{1}=3.7 m/s^{2}$

b) $a_{2}=3.68 m/s^{2}$

Explanation:

a) The displacement of the first object is 22.5 m, so we can use the next equation:

$v_{f}^{2}=v_{i}^{2}+2a\Delta x$

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.5}$

$a_{1}=3.7 m/s^{2}$

positive acceleration.

b) Using the same equation we can find the second value of the acceleration:

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.6}$

$a_{2}=3.68 m/s^{2}$

positive acceleration.

I hope it helps you!

8 0

4 years ago

Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900

Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron $m_{e}=9.11\times10^{-31}\ kg$

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}$

$E=0.582\ Mev$

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}$

$E=0.350\ Mev$

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0

3 years ago

There are three long parallel wires arranged so that, in cross-section, they occupy the points of an equilateral triangle. Is th

Harrizon [31]

Answer:

Yes, there is such a way.

Explanation:

If currents flow in the same direction in two or more long parallel wires, there will be an attractive force between the wires. If the current flows in different directions, there will be a repulsive force between the wires. In this case, these three parallel wires, can be be made to carry current in the same direction, creating an attractive force between all three wires.

Note that it is not possible to have at the least one of them carry current in the opposite direction and still have an attractive current between them.

8 0

3 years ago