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3 years ago

What is NOT a principle of genetics?

Physics

2 answers:

labwork [276]3 years ago

6 0

I don't know sorry ugly ahhhhhhhjhh

timurjin [86]3 years ago

3 0

The recessive trait will always show up

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A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude

Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$ .

Now, let us call $\lambda$ the charge per unit length, then we know that

$dQ = \lambda Rd\theta$ ;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda }{R}.$

Now, we know that

$\lambda = 3.0*10^{-9}C/m,$

$k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},$

and the radius of the semicircle is

$\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };$

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9}) }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0

3 years ago

Three long, straight wires are carrying currents that have the same magnitude. In C the current is opposite to that in A and B.

Nadusha1986 [10]

Answer:

(b) B

Explanation:

The direction of force on a current carrying wire in a magnetic field can be found using the right hand rule, which states that-"stretch the thumb in the direction of the current, and point the fingers in the direction of magnetic field. The direction of palm will then give the direction of force on the wire

On wire B the forces due to A and C act in the same direction and so strengthen each other. they get added up because the forces act in the same direction.

on wires A and C the forces (due to B and C and A and B

respectively) act in opposite directions and therefore tend to cancel out.

5 0

3 years ago

What is the magnitude of the electric field at a distance of 89 cm from a 27 μC charge, in units of N/C?

GuDViN [60]

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

$E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}$