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Afina-wow [57]
3 years ago
11

Write a paragraph that explains how one gene can be responsible for many traits, and how how many genes can be responsible for o

ne trait
Physics
1 answer:
gayaneshka [121]3 years ago
5 0
A gene is made of up several DNA's and these DNA's can be be responsible for many traits. There are also DNA's that contain the information on how the cells works and not on the traits of an organism. Therefore, one gene can have several DNA's that contain information about traits and another gene can have only a single DNA that contains information about a trait.
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What is the velocity of a wave that has a wavelength of 20 meters and frequency of 0.5<br> Hz?
k0ka [10]

Answer:

v = 10 m/s

Explanation:

recall that velocity is related to wavelength and frequency by the formula

v = fλ

where v = velocity, f = frequency and λ= wavelength

Simply substitute these into the formula:

v = fλ

v = (0.5)(20)

v = 10 m/s

3 0
3 years ago
Read 2 more answers
What kind of reaction occurs when potassium is placed in water? a. a single-displacement reaction c. a decomposition reaction b.
satela [25.4K]
The equation for potassium in water is:
K(s) + H20(l) --> H2(g) + K20(aq)
since a element and a compound are reacting, this is a single replacement reaction - which is a)
7 0
4 years ago
A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.
gavmur [86]

Answer:

(a) E = -1.02 \times 10^5~N/C

(b) E = -9.7 \times 10^4~N/C

Explanation:

(a) The electric field for a point charge is given by the following formula:

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

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}

Hence, dE is now:

dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}

Therefore, vertical component of dE now becomes

dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z =  \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C

(b) We will have a similar approach, but a simpler integral.

dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta  = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi

E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0
3 years ago
Find the length of a microscope having magnifying power 100, if the focal lengths of the objective and eye piece are 2 cm and 5
sp2606 [1]

Answer:

   L = 40 cm

Explanation:

A microscope is an optical instrument built by two lenses in such a way that the image of the first is formed within the distance of the other (eyepiece), so that the latter creates an enlarged virtual image of the object, for which the magnification of the microscope is the same to the multiplication of the magnification of each lens

    M = - L / f₀ (25 cm /f_{e})

where fo and fe are the focal lengths of the objective and eyepiece, 25 cm is the near vision distance and L is the length of the microscope

     L = - M f_{o} f_{e} / 25

let's calculate

     L = - (-100) 2 5/25

      L = 40 cm

7 0
3 years ago
Is winter eye grass autotrophic
inessss [21]
Yes I do believe it is
6 0
4 years ago
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