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Naya [18.7K]
3 years ago
14

What substance is the genetic material of living things?

Physics
1 answer:
Alik [6]3 years ago
3 0
I believe it's DNA.
Here're some sentences to explain:

Genetic material, also known as deoxyribonucleic acid<span> (</span>DNA<span>) and </span>Ribonucleic acid<span> (</span>RNA<span>), plays a fundamental role in the composition of living organisms.</span>
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¿Porque los sensores son importantes en nuestra vida diaria?
7nadin3 [17]
En la medicina y la biotecnología, los sensores son herramientas que detectan procesos biológicos, químicos, o físicos y luego transmiten o reportan esta información. Algunos sensores trabajan fuera del cuerpo, mientras que otros están diseñados para ser implantados dentro del cuerpo.

Algunos dispositivos de monitoreo constan de múltiples sensores que miden una serie de parámetros físicos o biológicos. Otros dispositivos pueden ser multifuncionales, incorporando sensores y luego suministrando un fármaco o intervención en base a la información obtenida de los sensores. Los sensores pueden ser también componentes en sistemas que procesan muestras clínicas, tales como los dispositivos cada vez más comunes conocidos como “lab-on-a-chip”.

Los sensores ayudan a los proveedores del cuidado de la salud y a los pacientes a monitorear las condiciones de la salud y asegurar que puedan tomar decisiones informadas sobre el tratamiento. Los sensores también se utilizan a menudo para monitorear la seguridad de los medicamentos, los alimentos, las condiciones ambientales, y otras sustancias que podríamos encontrar.
8 0
4 years ago
(a) Suppose that a NASCAR race car is moving to the right with a constant velocity of +93 m/s. What is the average acceleration
Doss [256]
A. the average acceleration would be 0 since there is no change in velocity

b. Average acceleration = change in velocity/time

93- (-93)/12

= 186 /12

= 15.5 m

hope this helps

4 0
3 years ago
1. What different types of shots are taken on the basketball court?
Lynna [10]

Answer:

Here are a few commonly used types of shooting in basketball.

  • Jump Shot. A jump shot is most frequently used for a mid to long-range shots, including shooting beyond the arc. ...
  • Hook Shot. A hook shot is when the shot is made while your body is not directly facing the basket. ...
  • Bank Shot. ...
  • Free Throw. ...
  • Layup. ...
  • Slam Dunk.
8 0
3 years ago
What minimum heat is needed to bring 250 g of water at 20 ∘C to the boiling point and completely boil it away? The specific heat
Amiraneli [1.4K]

Answer:633.8 KJ

Explanation:

Given

mass of water\left ( m\right )=250gm

Initial temperature\left ( T_i\right )=20^{\circ}C

Final temperature \left ( T_f\right )=100^{\circ}C

Specific heat of water \left ( c \right )=4190 J/kg-k

heat of vaporization\left ( L\right )=22.6\times 10^5 J/kg

Heat required for process\left ( Q\right )=heat to raise water temperature from 20 to 100 +Heat to vapourize water completely

Q=mc\left ( T_f-T_i\right )+mL

Q=0.25\times 4190\times \left ( 100-20\right )+0.25\times 22\times 10^5

Q=\left ( 0.838+5.5\right )\times 10^5

Q=6.338\times 10^5J=633.8 KJ

4 0
4 years ago
A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10°C. Two metallic blocks
ipn [44]

Answer:

A. 1,950 J/kgºC

Explanation:

Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the copper and the unknown metal.

The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:

Q = c * m* Δt

where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.

In our case, we can write the following equality:

(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (ccu*mcu*Δtcu) + (cₓ*mₓ*Δtₓ)

Replacing by the givens , and taking ccu = 0.385 J/gºC and cAl = 0.9 J/gºC, we have:

Qg= 0.9 J/gºC*100g*10ºC + 4.186 J/gºC*250g*10ºC  = 11,365 J(1)

Ql = 0.385 J/gºC*50g*55ºC + cₓ*66g*80ºC = 1,058.75 J + cx*66g*80ºC (2)

Based on all the previous assumptions, we have:

Qg = Ql

So, we can solve for cx, as follows:

cx = (11,365 J - 1,058.75 J) / 66g*80ºC = 1.95 J/gºC (3)

Expressing (3) in J/kgºC:

1.95 J/gºC * (1,000g/1 kg) = 1,950 J/kgºC

3 0
3 years ago
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