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frozen [14]
2 years ago
6

According to the Big Bang Theory, how long ago did the universe expand explosively into existence

Physics
2 answers:
Leni [432]2 years ago
5 0

According to some of the same evidence that is explained by the "Big Bang" theory, the universe expanded into existence roughly 13.7 billion years ago.

The Big Bang theory says nothing about the conditions at exactly the instant of the "beginning" or less than about 10⁻⁴³ second after it, and nothing about any "before".  

In fact, it says that anything "before" is "unknowable", and in some sense, that 'Time' itself started with the 'big bang'.  

Dmitry_Shevchenko [17]2 years ago
3 0

About 13.7 billion years ago

The Big Bang Theory states that the universe started about 13.7 billion years ago, and before that, everything was in 1 singularity.



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Assume that a gravitational anomaly in the solar system has shifted a field of asteroids into Earth’s orbit, and the field is no
Mandarinka [93]

Answer:

An asteroid is a minor planet of the inner Solar System. Historically, these terms have been applied to any astronomical object orbiting the Sun.

7 0
2 years ago
If the Earth were compressed in such a way that its mass remained the same, but the distance around the equator were just one-ha
vovikov84 [41]
If the distance around the equator is reduced by half, then the radius is also reduced by half.

Since the acceleration due to gravity is proportional to 1/(radius²),
the acceleration changes by a factor of 1/(1/2)² = 1/(1/4) = <em>4 </em>.

The acceleration due to gravity ... and also the weight of everything on Earth ...
becomes <em>4 times what it is now</em>.
6 0
3 years ago
Read 2 more answers
A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___
34kurt

Answer:

The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8·\hat j, in vector form

2) 1.0 km East = 1.0·\hat i, in vector form

3) 1.6 km South = -1.6·\hat j, in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8·\hat j + 1.0·\hat i +(-1.6·\hat j) = 1.2·\hat j + 1.0·\hat i

Final displacement, = 1.0·\hat i + 1.2·\hat j

Where;

Δx = The displacement in the x-direction = 1.0·\hat i

Δy = The displacement in the y-direction = 1.2·\hat j

The magnitude of the resultant displacement vector is given as follows

\left | d \right | = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0
3 years ago
Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl
lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

   m g sin\theta - F_{rope} = ma

 velocity is constant, a = 0

   m g sin\theta - F_{rope} =0

   F_{rope} = m g sin\theta

   F_{rope} = 70.1\times 9.8\times sin 8.6^0

  F_{rope}= 102.73\ N

b) now, when acceleration, a = 0.135 m/s²

   F_{rope}-m g sin\theta = ma

 velocity is constant, a = 0.135 m/s₂

   F_{rope} = m g sin\theta+ma

   F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135

  F_{rope}= 112.19\ N

7 0
3 years ago
PLS ANSWER DUE LATER TODAY !!!
Assoli18 [71]
Magna Carta. I’m pretty sure it’s the correct answer.
4 0
3 years ago
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