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3 years ago

What controls the traits an organism has?

1 answer:

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The Traits an organism displays are ultimately determined by the genes it inherited from its parents, in other words by its genotype. Variant copies of a gene are called alleles, and an individuals genotype is the sum of all the alleles inherited from the parents.

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A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

r < a:

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

a < r < b:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

r = b:

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

b < r < c:

E = 0

r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

r < c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

r<a:

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

At r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

At a < r < b:

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

$E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = b:

$E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

Again, the minus sign indicates the direction of the field towards the center.

At b < r < c:

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

At r < c:

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.5 cm/s on an otherwise placid lake.

nata0808 [166]

Find the given attachment

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3 years ago

Does the law of conservation of energy apply to waves?

dimulka [17.4K]

Answer:

Explanation:

Energy, as we have noted, is conserved, making it one of the most important physical quantities in nature. The law of conservation of energy can be stated as follows: Total energy is constant in any process. It may change in form or be transferred from one system to another, but the total remains the same.

4 0

3 years ago

En una librería un libro de 0.6 kg colocado inicialmente en un estante a 40 cm del suelo fue movido a otro estante con una altur

jeyben [28]

Considerando la definición de energía potencial:

la energía potencial del libro inicialmente es 2,3544 J y luego de ser movido a otro estante es 7,6518 J.
el cambio en la energía potencial del libro es 5,2974 J.

<h3 /><h3>Definición de energía potencial</h3>

La energía potencial es la energía que mide la capacidad que tiene un sistema para realizar un trabajo en función de su posición. En otras palabras, esta es la energía que tiene un cuerpo situado a una determinada altura sobre el suelo.

La energía potencial gravitatoria es la energía asociada con la fuerza gravitatoria. Esta dependerá de la altura relativa de un objeto a algún punto de referencia, la masa, y la fuerza de la gravedad.

Entonces para un objeto con masa m, en la altura h, la expresión aplicada a la energía gravitacional del objeto es:

Ep= m×g×h

Donde

Ep es la energía potencial en julios (J).
m es la masa en kilogramos (kg).
h es la altura en metros (m).
g es la aceleración de caída en m/s² (aproximadamente 9,81 m/s²).

<h3>Energía potencial en cada lugar</h3>

En primer lugar, se sabe inicialmente del libro:

m= 0,6 kg
g= 9,81 m/s²
h= 40 cm= 0,4 m (siendo 1 cm= 0,01 m)

Entonces, reemplazando en la definición de energía potencial:

Ep1= 0,6 kg× 9,81 m/s²× 0,4 m

Resolviendo:

Ep1= 2,3544 J

Por otro lado, se sabe los siguientes datos del libro luego de ser movido:

m= 0,6 kg
g= 9,81 m/s²
h= 1,30 m

Entonces, reemplazando en la definición de energía potencial:

Ep2= 0,6 kg× 9,81 m/s²× 1,30 m

Resolviendo:

Ep2= 7,6518 J

En resumen, la energía potencial del libro inicialmente es 2,3544 J y luego de ser movido a otro estante es 7,6518 J.

<h3>Cambio de energía potencial</h3>

El cambio en la energía potencial del libro será la diferencia entre la energía potencial luego de ser movido y la energía potencial inicial del libro:

ΔEp= Ep2 - Ep1

ΔEp= 7,6518 J - 2,3544 J

ΔEp= 5,2974 J

Finalmente, el cambio en la energía potencial del libro es 5,2974 J.

Aprende más sobre energía potencial:

brainly.com/question/21370800

brainly.com/question/25163940

brainly.com/question/25960490

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2 years ago

How is resonance used in musical instruments

sweet-ann [11.9K]