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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 wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit

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A) $+1.67 rad/s^2$

The angular acceleration of the wheel is given by

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

$\omega_f = 4.00 rad/s$ is the final angular velocity (anticlockwise, so with a positive sign)

$\Delta t= 6.00 s - 0=6.00 s$ is the time interval

Substituting into the equation, we find the angular acceleration:

$\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2$

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant $t_1$ where the angular velocity becomes positive (so, $\omega_i=0$ ) and the time corresponding to the final instant $t_2 = 6.0 s$ , where $\omega_f = +6.00 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s$

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant $t_1=0$ when $\omega_i=-4.00 rad/s$ ) and the time corresponding to the instant in which the velovity becomes positive $t_2$ , when $\omega_f = 0 rad/s$ . We can find this time interval by using