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frozen [14]
3 years ago
6

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

Physics
2 answers:
Leni [432]3 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]3 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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Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. Exp
VARVARA [1.3K]

The unbalanced reaction is

<em>a</em> NH₄NO₃   ⇒   <em>b</em> N₂ + <em>c</em> O₂ + <em>d</em> H₂O

where <em>a</em>, <em>b</em>, <em>c</em>, and <em>d</em> are unknown constants.

Count how many times each element appears on either side of the reaction.

• reactants:

N = 2<em>a</em>, H = 4<em>a</em>, O = 3<em>a</em>

• products

N = 2<em>b</em>, O = 2<em>c</em> + <em>d</em>, H = 2<em>d</em>

<em />

Now we solve the system of equations,

2<em>a</em> = 2<em>b</em> … … … [1]

4<em>a</em> = 2<em>d</em> … … … [2]

3<em>a</em> = 2<em>c</em> + <em>d</em> … … … [3]

<em />

From [1] we immediately have

<em>a</em> = <em>b</em>

In [2], we get

2<em>a</em> = <em>d</em>

and substituting for <em>d</em> in [3] gives

3<em>a</em> = 2<em>c</em> + 2<em>a</em>

<em>a</em> = 2<em>c</em>

<em />

Let <em>c</em> = 1; then

• <em>a</em> = 2×1 = 2

• <em>b</em> = 2

• <em>d</em> = 2×2 = 4

So, the balanced reaction is

2 NH₄NO₃   ⇒   2 N₂ + O₂ + 4 H₂O

8 0
3 years ago
A 2.5 m -long wire carries a current of 8.0 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the
leva [86]

Answer:

Explanation:

Let the magnetic field be B = B₁i + B₂j + B₃k

Force = I ( L x B )  , I is current , L is length and B is magnetic field .

In the first case

force = - 2.3 j N

L = 2.5 i

puting the values in the equation above

- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]

= - 20 B₃ j + 20 B₂ k

comparing LHS and RHS ,

20B₃ = 2.3

B₃ = .115

B₂ = 0

In the second case

L = 2.5 j

Force = I ( L x B )

2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )

=  - 20 B₁ k + 20B₃ i

2.3i−5.6k = - 20 B₁ k + 20B₃ i

B₃ = .115

B₁ = .28

So magnetic field B = .28 i + .115 B₃

Part A

x component of B = .28 T

Part B

y component of B = 0

Part C

z component of B = .115 T .

8 0
3 years ago
Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s
alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

  • Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
  • Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
  • vₓ₀ = v * cos θ (1)
  • where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
  • Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        x_{f} = v_{ox} * t = v_{o} * cos \theta * t  (2)

  • Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)

  • ⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

  • At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
  • Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
  • vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

  • At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

  • Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       \Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)

  • Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
  • v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
  • Replacing (7) in (6), we get:

       \Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)

e)

  • When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
  • The horizontal component, since it keeps constant, is just v₀x:
  • v₀ₓ = 13.7 m/s
  • The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       v_{fy} = v_{oy} - g*t  (9)

  • Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s  (10)

  • Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)

3 0
3 years ago
Why does the moon exert a greater tidal influence than the sun?
andreyandreev [35.5K]
Well, I'm not sure right now that it actually does.

But if it does, that's because the sun is about 400 times
FARTHER from the Earth than the moon is.
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3 years ago
Dry air will break down if the electric field exceeds about 3.0×106v/m. part a what amount of charge can be placed on a capacito
Genrish500 [490]
The solution for this problem is:The charge would be now equal to:(electric constant) multiplied by the (field strength) multiplied by the (area) so plugging in our values, will give us:8.85 * 10^-12 As / (V * m) * 3 * 10^6 V/m * 0.055 m^2 = 1.46 e-6 amperes would be the answer
7 0
3 years ago
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