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3 years ago

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The big bang theory of the information and expansion of the universe is supported by the observed

1 answer:

dusya [7]3 years ago

6 0

By Hubble theory in which universe is expanding,

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What is the only bone of the skull that moves?

Darina [25.2K]

Answer:

Mandible.

Explanation:

The mandible in the bone that allows you to open and close your mouth, otherwise known ad the jawbone.

3 0

3 years ago

Read 2 more answers

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

The object will feel minimum force of gravity at the

Alika [10]

Answer:

space , small amount of gravity is found in the space ,infact we can say that there is no gravity in the space

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3 years ago

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What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s

AlexFokin [52]

Answer:

$K=0.023J$

Explanation:

From the question we are told that:

Mass $m=0.15$

Velocity $v=0.5m/s$

Angular Velocity $\omega=8.4rad/s$

Generally the equation for Kinetic Energy is mathematically given by

$K=\frac{1}{2}M(v^2+\frac{1}{2}R^2\omega^2)$

$K=\frac{1}{2}0.15(0.5^2+\frac{1}{2}(0.038)^2.(8.4rad/s^2))$

$K=0.023J$

8 0

3 years ago

What is NOT a raw minteral used in photosyntheisis

Yuliya22 [10]

The photosynthesis equation is :

H2O + CO2 + sunlight (energy) > C6H12O6 + O2

So water (H2O) and carbon dioxide (CO2) are the raw materials need for photosynthesis.

6 0

3 years ago