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2 years ago

Which occurrence would contradict the big bang theory?

Physics

2 answers:

maw [93]2 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]2 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 an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. N

galina1969 [7]

Answer:

The temperature of the cooler substance was close to the room temperature. Therefore, the system experienced less change

Explanation:

Generally, in a closed system containing two bodies at different temperatures, there is a flow of heat energy from the body at a higher temperature to the body at a lower temperature. The effect is more significant when there is a large temperature difference between the bodies. However, if the temperature difference is small or insignificant, the change will be less.

3 0

2 years ago

A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it

Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid $SG$ (also called relative density) is the ratio of the density of that solid $\rho_{rock}$ to the density of water $\rho_{water}=1 kg/m^{3}$ (normally at $4\°C$ ):

$SG=\frac{\rho_{rock}}{\rho_{water}}$ (1)

On the other hand, the density of the rock is calculated by:

$\rho_{rock}=\frac{m_{rock}}{V_{rock}}$ (2)

Where:

$m_{rock}$ is the mass of the rock

$V_{rock}=\frac{4}{3} \pi r^{3}$ is the volume of the rock, since is spherical

Well, we already know the value of $\rho_{water}$ , but we need to find $\rho_{rock}$ in order to find the rock's specific gravity; and in order to do this, we firsly have to find $m_{rock}$ and then calculate $V_{rock}$ :

In the case of the mass of the rock, we can calclate it by the following equation:

$W_{rock}=m_{rock}g_{mars}$ (3)

Where:

$W_{rock}$ is the weight if the rock in mars

$g_{mars}=3.7 m/s^{2}$ is the acceleration due gravity in Mars

Isolating $m_{rock}$ :

$m_{rock}=\frac{W_{rock}}{g_{mars}}$ (4)

$m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}$ (5)

To find $W_{rock}$ we can use the following equation of the potential gravitational energy $U$ :

$U=W_{rock}H$ (6)

Where:

$U=2 J=2 Nm$ is the potential energy

$H=20 cm \frac{1m}{100 cm}=0.2 m$ is the height at which the rock has the mentioned potential energy

Isolating $W_{rock}$ :

$W_{rock}=\frac{U}{H}$ (7)

$W_{rock}=\frac{2 Nm}{0.2 m}$ (8)

$W_{rock}=10 N$ (9)

Substituting (9) in (5):

$m_{rock}=\frac{10 N}{3.7 m/s^{2}}$ (10)

$m_{rock}=2.702 kg$ (11)

Substituting (11) in (2):

$\rho_{rock}=\frac{2.702 kg}{V_{rock}}$ (12) At this point we only need to find the volume of the rock, knowing its diameter is $d=10 cm$ , hence its radius is $r=\frac{d}{2}=5 cm$

$V_{rock}=\frac{4}{3} \pi (5 cm)^{3}$ (13)

$V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3}$ (14)

Substituting (14) in (12):

$\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}}$ (15)

$\rho_{rock}=5166.34 kg/m^{3}$ (16)

Substituting (16) in (1):

$SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}}$ (17)

Finally we obtain the specific gravity of the rock:

$SG=5166.347$

7 0

2 years ago

An object is released from rest near and above Earth’s surface from a distance of 10m. After applying the appropriate kinematic

Damm [24]

Answer:

v_y = 12.54 m/s

Explanation:

Given:

- Initial vertical distance y_o = 10 m

- Initial velocity v_y,o = 0 m/s

- The acceleration of object in air = a_y

- The actual time taken to reach ground t = 3.2 s

Find:

- Determine the actual speed of the object when it reaches the ground?

Solution:

- Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:

y = y_o + v_y,o*t + 0.5*a_y*t^2

0 = 10 + 0 + 0.5*a_y*(3.2)^2

a_y = - 20 / (3.2)^2 = 1.953125 m/s^2

- Use the principle of conservation of total energy of system:

E_p - W_f = E_k

Where, E_p = m*g*y_o

W_f = m*a_y*(y_i - y_f) ..... Effects of air resistance

E_k = 0.5*m*v_y^2

Hence, m*g*y_o - m*a_y*(y_i - y_f) = 0.5*m*v_y^2

g*(10) - (1.953125)*(10) = 0.5*v_y^2

v_y = sqrt (157.1375)

v_y = 12.54 m/s

4 0

2 years ago

As planets move toward the sun, they tend to: A. move at constant speed B. speed up C. burn up D. slow down

dolphi86 [110]

Answer:

B. speed up

Explanation:

The acceleration of an object due to a body's gravity is:

g = GM / r²

where G is the universal constant of gravitation,

M is the mass of the body,

and r is the distance from the body.

As a planet approaches the sun, r decreases. As r decreases, g increases.

8 0

2 years ago

Jenny and Alyssa are members of the cross-country team. On a training run, Jenny starts off and runs at a constant 3.8 m/s. Alys

guapka [62]

Answer:

285 seconds

Explanation:

Jenny speed is 3.8 m/s

Alyssa speed in 4.0 m/s

Alyssa starts after 15 seconds

Find the distance covered by Jenny, when Alyssa starts

Distance=Speed*time

Distance covered by Jenny in 15 seconds= 3.8×15=57m

Relative speed of the two members heading same direction will be;

4.0m/s-3.8m/s=0.2m/s

To find the time Alyssa catch up with Jenny you divide the distance to be covered by Alyssa by the relative speed of the two

Distance=57m, relative speed=0.2m/s t=57/0.2 =285 seconds

=4.75 minutes

5 0

2 years ago

Read 2 more answers