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alexdok [17]
3 years ago
8

Do you think scientists will ever be able to recreate a living creature from the distant past successfully?

Physics
2 answers:
Agata [3.3K]3 years ago
6 0
It probably isn't likely
lilavasa [31]3 years ago
3 0
Are we talking about extinct?

I think it could be possible. However, the creature may not last long. Creatures from the distant past died off for a reason, they couldn't survive. So, with the sudden change in environment, they won't last long, especially if they didn't last in the past.
You might be interested in
What did the asymptote say to the removable discontinuity worksheet answers?
ra1l [238]

“Don't hand that holier than thou line to me” is what the asymptote said to the removable discontinuity.

 

 

The distance between the curve and the line where it approaches zero as they tend to infinity is the line in the asymptote of a curve. This is unusual for modern authors but in some sources the requirement that the curve may not cross the line infinitely often is included.

 

The point that does not fit the rest of the graph or is undefined is called a removable discontinuity. By filling in a single point, the removable discontinuity can be made connected.

6 0
3 years ago
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
2 years ago
what will happen to the brightness of the other bulbs in a series circuit if any of the bulbs were to burn out?
Ivenika [448]

Answer:

not work.

Explanation:

if u are saying in a series circuit...

if 1 build burns out and theres other bulbs the circuit wont work anymore.

6 0
3 years ago
In which of the two situations described is more energy transferred?
Furkat [3]

Answer:

More energy is transferred in situation A

Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m·H_f

Where;

H_f = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0
2 years ago
studies show that the ocean surface temperature is increasing which of the following is a possible consequences
katen-ka-za [31]
One possible consequence is that the warmer temperature cause the polar ice to melt even faster
6 0
3 years ago
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