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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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In Thomson’s experiment, why was the glowing beam repelled by a negatively charged plate?

Svetllana [295]

The glowing beam was repelled by a negatively charged plate because they were negatively charged

<h3>What are the nature of charges?</h3>

The nature of charges refers to the properties of charges.

There are two types of charges:

negative charges
positive charges

The law of electricity states that opposite charges attract whereas like charges repel.

Therefor, in Thomson’s experiment, the glowing beam was repelled by a negatively charged plate because they were negatively charged

In conclusion, like charges repel while opposite charges attract.

Learn more about charges at: brainly.com/question/12781208

#SPJ1

5 0

2 years ago

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas

Pavel [41]

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

3 0

3 years ago

For a particular type of motion, the velocity is zero but the speed is a nonzero quantity. Which statement can you make about th

masha68 [24]

Answer:

because speed is the modulus of velocity which is a vector

the velocity to be zero it must be a round trip

Explanation:

This is because speed is the modulus of velocity which is a vector.

For the velocity to be zero it must be a round trip, therefore the resulting vector zero

On the other hand, the speed of the module is the same in both directions

8 0

3 years ago

A point charge q1=2.0μC is located on the positive y axis at y=0.30m, and an identical charge q2 is at the origin. Find the magn

dalvyx [7]

Answer:

(A) 0.279N at angle 38.02°

(B) 0.701N

Explanation:

(A) The net force on q3 is given as:

F = Fxi + Fyj

Fx is the x component of the force

Fy is the y component of the force

Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0

Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx

First let us find y and angle x from the diagram.

Using Pythagoras theorem,

y² = 0.3² + 0.4²

y² = 0.25

y = 0.5m

Using SOHCAHTOA to find x,

sinx = 0.4/0.5

x = 53.13°

Electrostatic force, F is given as:

F = kqQ/r²

Where k = Coulumbs constant

F(1,3) = (k*q1*q3) / r²

F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)

F(1, 3) = 0.288N

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = 0.45N

Therefore,

Fx = -0.288cos36.87 + 0.45

Fx = 0.22N

Fy = 0.288cos53.13

Fy = 0.172N

=> F = 0.22i + 0.172j

The magnitude of the force will be

F(mag) = √(0.22² + 0.172²)

F(mag) = 0.279N

The direction of the force makes will be

tanθ = Fy/Fx

tanθ = 0.172/0.22 = 0.781

θ = 38.02° to the x axis.

(B) q2 = - 2.0 * 10^(-6)

This implies that:

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = -0.45N

Therefore,

Fx = -0.288cos36.87 - 0.45

Fx = -0.68N

Fy = 0.172N

=> F = - 0.68i + 0.172j

The magnitude of the force will be

F(mag) = √((-0.68)² + 0.172²)

F(mag) = 0.701N

tanθ = 0.172/0.68

θ = 14.19° to the x axis

4 0

3 years ago

Read 2 more answers

Two resistors, R1 and R2, are

Aleonysh [2.5K]

Answer:

The value of R₂ is equal to 24.75 ohms.

Explanation:

Given that,

Two resistors, R₁ and R₂, are connected in parallel.

The equivalent resistance is 14.5 ohms

We need to find the value of R₂.

When two resistors are connected in parallel. The equivalent resistance is given by :

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{14.5}=\dfrac{1}{35}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_2}=\dfrac{1}{14.5}-\dfrac{1}{35}\\\\\dfrac{1}{R_2}=0.04039\\\\R_2=\dfrac{1}{0.04039}\\\\R_2=24.75\ \Omega$

So, the value of R₂ is equal to 24.75 ohms.

8 0

3 years ago