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MrMuchimi
2 years ago
5

If the Universe was born at the Big Bang, what existed before then?​

Physics
1 answer:
Aleonysh [2.5K]2 years ago
8 0
<h2>Answer:</h2>

<em>Hey, </em>

<h3><u>QUESTION)</u></h3>

This is a complicated question that would be debated by many physicists. How to answer this question when we are struggling with the genesis of the universe? Many theories on the origin of the universe have been put forward, but the Bing Bang is probably, or at least to date, the most likely theory. If today the universe is expanding, some astronomers hypothesise that before the Bing Bang the universe already existed but that it was condensing, according to this hypothesis the Bing Bang would correspond to the limit point of compression of the universe.  

But if we remain in the traditional theory the Bing Bang is at the origin of all the matter we know, as well as light. Before the Bing Bang, in other words before the universe, there was NOTHING.

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Find the torque required for the shaft to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 re
Luba_88 [7]

Answer:

(a) 152.85 Nm

(b) 1528.5 Nm

Explanation:

According to the formula of power

P = τ ω

ω = 2 π f

(a) f = 2500 rpm = 2500 / 60 = 41.67 rps

So, 40 x 1000 = τ x 2 x 3.14 x 41.67

τ = 152.85 Nm

(b) f = 250 rpm = 250 / 60 = 4.167 rps

So, 40 x 1000 = τ x 2 x 3.14 x 4.167

τ = 1528.5 Nm

3 0
3 years ago
How tall would a tower need to be if the period of a pendulum were 30. seconds
Virty [35]
We use the following expression

T = 2*pi *sqrt(l/g)

Where T is the period of the pendulum

l is the length of the pendulum

and g the acceleration of gravity

We solve for l

l = [T/2*pi]² *g = [30s/2*pi]²* 9.8 [m/s²] = 223.413 m

The tower would need to be at least 223.413 m high
4 0
3 years ago
Einstein calculated that ripples of gravity travel at exactly the speed of _____
damaskus [11]

Answer:

299,792,458 m/s = speed of light

Explanation:

6 0
3 years ago
T/F: Stars die.<br> True<br> False
Yuki888 [10]

Answer:

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8 0
3 years ago
2. Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 x 10-9 C and the oth
Maru [420]

Answer:

A. -2.16 * 10^(-5) N

B. 9 * 10^(-7) N

Explanation:

Parameters given:

Distance between their centres, r = 0.3 m

Charge in first sphere, Q1 = 12 * 10^(-9) C

Charge in second sphere, Q2 = -18 * 10^(-9) C

A. Electrostatic force exerted on one sphere by the other is:

F = (k * Q1 * Q2) / r²

F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²

F = -2.16 * 10^(-5) N

B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:

Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))

= - 6 * 10^(-9) C

Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C

Hence the electrostatic force between them is:

F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²

F = 9 * 10^(-7) N

7 0
3 years ago
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