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vladimir1956 [14]
3 years ago
11

The base for hair follicles is in the _____. epidermis fatty layer dermis dentin

Physics
1 answer:
Elis [28]3 years ago
8 0

Answer:

dermis

Explanation:

epidermis

dermis

hypodermis

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I need help the option for D is the golfball isnt accelerating
algol13

Answer:

es b

Explanation:

no se

8 0
3 years ago
A thin rod of length L and total charge Q has the nonuniform linear charge distribution λ(x)=λ0x/L, where x is measured from the
marissa [1.9K]

Answer:

Explanation:

λ(x) = λo x/ L

(a) The total charge is Q.

Q=\int_{0}^{L}dq

Q=\int_{0}^{L}\frac{\lambda _{0}x}{L}dx

Q=\frac{\lambda _{0}}{2L}\left ( x^{2} \right )_{0}^{L}

Q=\frac{\lambda _{0}}{2L}

λo = 2Q/L

(b)

Let at a distance x from the origin the charge is dq.

so, dq = (2Q/L) x/ L dx

dq=\frac{2Qx}{L^{2}}dx

The potential due to this small charge at a distance d to the left of origin

dV = \frac{KdQ}{d+x}

\int_{0}^{V}dV = \frac{2KQ}{L^{2}}\int_{0}^{L}\frac{xdx}{d+x}

V = \frac{2KQ}{L^{2}}\int_{0}^{L}\left ( 1- \frac{d}{d+x}\right )dx

V = \frac{2KQ}{L^{2}}\times \left ( x-dln(d+x) \right )\int_{0}^{L}

V = \frac{2KQ}{L^{2}}\times \left ( L-dln(d+L)-0+dlnd \right )

V = \frac{Q}{4\pi \epsilon _{0}L^{2}}\times \left ( L+d\times ln\left (\frac{d}{d+L}  \right )\right )

4 0
3 years ago
Before Newton and develop reflecting telescopes what instruments did astronomers use?
never [62]

The first telescopes were refracting telescopes ... a tube
with lenses at both ends.  The quality of the image depends
on the quality of both lenses, because the light must pass
through both of them. 

The reflecting telescope was an improvement, because the
light doesn't have to go through the mirror.  The quality of the
primary mirror only depends on the shape and shinyness of
its surface. 

If you could carve an optically perfect parabolic surface out of
a piece of wood and coat it with a thin layer of shiny silver, then
you could use a log to make a good reflecting telescope.
II seem to recall that Newton made one out of brass !
7 0
3 years ago
A lightweight plastic rod has a mass of 2.0 kg attached to one end and a mass of 2.5 kg attached to the other end. The rod has a
ipn [44]

The distance from the 2.0 kg mass at which a string should be attached to balance the rod is 0.67 m.


<h3>What is distance?</h3>

Distance can be defined as the horizontal length between two points.

To calculate the distance at which the string must be attached to balance the rod, we use the formula below.

Formula:

  • m₁gx = m₂g(1.2-x)................. Equation 1

From the question,

Given

  • m₁ = 2.0 kg
  • m₂ = 2.5 kg
  • x = Distance of the string from the 2.0 kg mass
  • g = Acceleration due to gravity.= 9.8 m/s²

Substitute these values into equation 1

  • 2(9.8)(x) = 2.5(9.8)(1.2-x)

Solve for x

  • 19.6x = 29.4-24.5x

Collect like terms

  • 19.6x+24.5x = 29.4
  • 44.1x = 29.4
  • x = 29.4/44.1
  • x = 0.67 m

Hence, the distance from the 2.0 kg mass at which a string should be attached to balance the rod is 0.67 m.

Learn more about distance here: brainly.com/question/17273444

5 0
2 years ago
The bending of light as it passes into a transparent material of different optical intensity is known as?
galina1969 [7]

Well, it's actually optical "density", but I understand
what you're getting at.

That phenomenon is called "refraction".

It happens with any wave motion.  It's very prominent in the
progress of seismic waves, radio waves, sound waves etc.

4 0
3 years ago
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