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3 years ago

Which occurrence would contradict the big bang theory?

Physics

2 answers:

maw [93]3 years ago

6 0

<h2>Answer:</h2>

<u>The statement that contradicts with the big bang theory is</u><u> Over time, distant galaxies move progressively closer to Earth.</u>

<u></u>

<h2>Explanation:</h2>

In 1927, the Belgian Catholic priest Georges Lemaître proposed an expanding model for the universe to explain the observed redshifts of spiral nebulae, and calculated the Hubble law. Hence he laid the foundation of Big Bang Theory according to which the universe is expanding. There is no such statement that says about the contraction of all heavenly bodies towards the earth.

OLga [1]3 years ago

3 0

The big bang theory states that all the matter that makes the universe was, at the beginning, I in a single infinitesimally small and infinitesimally dense point. Then, the point explodes given birth to the universe that we know. One of the major foundations that the universe was originated in that explosion called the big bang was made by Hubble when he proved that the universe is expanding; therefore, generally speaking galaxies are moving away from earth. The occurrence that contradicts the big bang theory is: “Over time, distant galaxies move progressively closer to Earth”

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A pumpkin is thrown horizontally off of a building at a speed of 2.5 m/s and travels a horizontal distance of 12 m before hittin

LUCKY_DIMON [66]

Answer:

-47.04

Explanation:

Divide 12 m by 2.5 m/s which equals 4.8s

And then use 4.8s to multiply it with -9.8

4.8(-9.8)= -47.04 m/s

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3 years ago

What is the distance between lines on a diffraction grating that produces a second-order maximum for 760-nm red light at an angl

Nataly [62]

The condition for maximum intensity is the same as that for a double slit. However, angular separation of the maxima is generally much greater because the slit spacing is so small for a diffraction grating. For a diffraction grating with lines/mm = lines/inch, the slit separation is d = micrometers = x10^ m.

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2 years ago

Why can’t we see the molecules moving in a solid or liquid?

harina [27]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

A missile is moving 1350 m/s at 25.0° angle. It needs to hit in a 55.0° direction in 10.20 s. What is the direction of its final

77julia77 [94]

Answer:

final velocity = 3504 m/s

Explanation:

<em>Given data:</em>

velocity of missile = Vi = 1350m/s

angle at which missile is moving = 25degree

distance between missile and targets = 23500m

angle between target and missile=55degree

time=10.2s

Final velocity: ?

<em>Formula:</em>

x = Vx*t + ½*ax*t²

Let x be the horizontal component of distance

x = ertical component of distance

t-time

ax = horizontal component of acceleration

ay = Vertical component of acceleration

Vx = horizontal component of velocity

Vy = Vertical component of velocity

<em>Solution:</em>

x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²

ax = 19.2 m/s²

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s

<em>similarly vertically:</em>

y = Vy*t + ½*ay*t²

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²

ay = 258 m/s²

V'y = Vy + ay*t

= 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s

V = √(V'x² + V'y²)

= 3504 m/s

8 0

3 years ago

A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150

vodomira [7]

Answer:

a) 607.5 J

b) 160.531875 J

c) 0 J

d) 0 J

e) 2.925 m\s

Explanation:

The given data :-

Mass of the box ( m ) = 37.5 kg.

Displacement made by box ( x ) = 4.05 m.

Horizontal force ( F ) = 150 N.
The co-efficient of friction between box and floor ( μ ) = 0.3
Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875

Solution:-

a) The work done by applied force ( W )

W = force applied × displacement = 150 × 4.05 = 607.5 J

b) The increase in internal energy in the box-floor system due to friction.

Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N

Change in internal energy = change in kinetic energy.

ΔU = ( K.E )₂ - ( K.E )₁

Since the initial velocity is zero so the ( K.E )₁ = 0

ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

c) The work done by the normal force .

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

d) The work done by the gravitational force.

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

e) The change in kinetic energy of the box

( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

f) The final speed of the box

( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²

v² = 8.56

v = 2.925 m\s.

5 0

4 years ago