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Nataly [62]
3 years ago
5

If the Sun were to collapse into a black hole, the point of no return for an investigator would be approximately 3 km from the c

enter singularity. Would the investigator be able to survive visiting even 300 km from the center?
Physics
1 answer:
Marysya12 [62]3 years ago
4 0

Answer:

1.97*10⁴N .He cannot survive.

Explanation:

The<em> </em><em>Tidal force</em> can be defined as the difference in force between center of  two bodies like center of earth and other bodies.This force stretches  the body too and fro from the center of mass of another body due to the difference in the gravitational force.

The tidal force is responsible for various phenomena like ocean tides,celestial body tearing up.

The tidal force is the effect of massive body gravitationally affecting another body.

The tidal force is the difference between the gravitational force of two bodies and directly proportional to the mass of body affecting another body and inversely proportional to the square of radius of the distance between center and the object.

Here let us consider height of person as 2m and d=1m

  ΔF=\frac{GMm}{(R-d)^{2} } -\frac{GMm}{(R+d)^{2} }

       =GMm[\frac{1}{(R-d)^{2} }-\frac{1}{(R+d)^{2} }]

         =(6.67*10⁻¹¹)(1.99*10³⁰)(1.0)[\frac{1}{(300*10^3-1)^{2} }-\frac{1}{(300*10^3+1)^{2} }}

   ΔF=1.97*10⁴N.

Thus he cannot survive

           

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It was progressively developed over time as the scientific knowledge advanced; for this reason many modifications and corrections might be done in the future.

Its usefulness lies in the fact that it allows the existing elements to be organized in a more structured and coherent way, according to the chemical properties they possess. Dividing the table into rows and columns, which represent the periods and groups or families.

Then, with the location and classification of an element according to its group, we can determine how it acts by knowing its chemical and physical characteristics.

This is how with this configuration can be distinguished 4 sets of chemical elements, according to the ease of their atoms to lose or gain electrons, transforming into ions: metals, semimetals, non-metals and noble gases.

This has helped to predict the existence of various elements that have not yet been discovered, because by elements already located in the table and the periodicity found, <u>there are still empty spaces that indicate the composition of the element that has not yet been found</u>.

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6 0
3 years ago
Renewable energy sources are ____________________?
Nina [5.8K]

Answer:

Eco-friendly

Explanation:

6 0
4 years ago
Explain the term inertia<br>​
andreyandreev [35.5K]

Answer: See explanation

Explanation:

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Inertia keeps objects and things in place and it holds the universe together. When there's no force that's acting in an object, such object will continue to move in a straight line and also at a constant speed.

8 0
3 years ago
ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami
yuradex [85]

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

Explanation :

The given balanced chemical reaction is,

2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{product}

\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]

\Delta H^o=62.2kJ=62200J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{product}

\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]

\Delta S^o=132.67J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 298 K.

\Delta G^o=(62200J)-(298K\times 132.67J/K)

\Delta G^o=22664.34J=22.66kJ

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

6 0
4 years ago
Which of the following would decrease in size during the contraction of a sarcomere? The width of the I-bands The width of the A
ANEK [815]

Hi!


The correct answer would be: the width of I-bands


The sacromere is the smallest contractile unit of striated muscles. These units comprise of filaments (fibrous proteins) that, upon muscle contraction or relaxation, slide past each other. The sacromere consists of thick filaments (myosin) and thin filaments (actin).


<em>Refer to the attached picture to clearly see the structure of a sacromere.</em>


<u>When a sacromere contracts, a series of changes take place which include:</u>

<em>- Shortening of I band, and consequently the H zone</em>

<em>- The A line remains unchanged</em>

<em>- Z lines come closer to each other (and this is due to the shortening of the I bands) </em>

The only changes that take place occur in the zones/areas in the sacromere (as mentioned), not in the filaments (actin and myosin) that make the up the sacromere; hence all other options are wrong.


Hope this helps!

8 0
3 years ago
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