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Katena32 [7]
3 years ago
9

The fraction of nonreflected radiation that is transmitted through a 5-mm thickness of a transparent material is 0.95. if the th

ickness is increased to 12 mm, what fraction of light is transmitted?

Physics
1 answer:
Zielflug [23.3K]3 years ago
7 0
Check the attached file for the answer.

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describe how light reflecting from a mirror can produce an image. In particular, explain how mirrors can produce images that are
VikaD [51]

Answer:

A larger image is produced when  d_i > d_o

A smaller image is produced when d_i < d_o

An upright image  is produced when m is positive

An upright image  is produced when m is negative

Explanation:

The mirror equation is given as follows;

\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}

m =-\dfrac{d_i}{d_o} = \dfrac{h_i}{h_o}

For concave mirrors, f = focal length

d_i = Image distance from the mirror (-ve d_i = Image is behind the mirror +ve d_i = Image is in front of the mirror)

d_o = Object distance from the mirror (-ve d_o = Object is behind the mirror +ve d_o = Object is in front of the mirror)

m = Magnification (-ve m = Inverted image +ve m = upright image)

h_i = Image height

h_o = Object height

f = Focal length of the mirror

To produce a larger image d_i > d_o

To produce a smaller image d_i < d_o

To produce an upright image, m should be positive hence, d_i will be negative or the image will appear behind the mirror

To produce an inverted image, m should be negative hence, d_i will be positive or the image will form in front of the mirror.

5 0
3 years ago
Which statement is true when acids and bases are mixed together
kap26 [50]

Answer:Acids and bases cannot mix together. Acids and bases will neutralize each other. Acids, but not bases, can change the pH of a solution. Acids donate hydroxide ions (OH–); bases donate hydrogen ions (H+)

8 0
3 years ago
A typical garden hose has an inner diameter of 5/8". Let's say you connect it to a faucet and the water comes out of the hose wi
castortr0y [4]
Since speed (v) is in ft/sec, let's convert our diameters from inches to feet:
1) 5/8in = 0.625in
0.625in × 1ft/12in = 0.0521ft
2) 0.25in × 1ft/12in = 0.021ft
Equation:
v = 4q \div ( {d}^{2} \pi) \: where \: q = flow \\ v = velocity \: (speed) \: and \:  \\ d = diameter \: of \: pipe \: or \: hose \\ and \: \pi = 3.142
we \: can \: only \: assume \:that \\  flow \: (q) \:stays \: same \: since \: it \\  isnt \: impeded \: by \:  anything \\ thus \:it  \: (q)\:  stays \: the \: same \:  \\ so \: 4q \: can \: be \: removed \: from \:  \\ the \: equation
then \: we \: can \: assume \: that \: only \\ v \: and \: d \: change \: leading \:us \: to >  >  \\ (v1 \times {d1}^{2} \pi) = (v2  \times   {d2}^{2}\pi)
both \: \pi \: will \: cancel \: each \: other \: out \:  \\ as \: constants \:since \: one \: is \: on \\ each \: side \: of \: the \:  =

(v1  \times   {d1}^{2}) = (v2 \ \times {d2}^{2}) \\ (7.0 \times   {0.052}^{2}) = (v2  \times   {0.021}^{2}) \\ divide \: both \: sides \: by \:  {0.021}^{2} \\ to \: solve \: for \: v2 >  >
v2 = (7.0)( {0.052}^{2} ) \div ( {0.021}^{2})  \\ v2 = (7.0)(.0027) \div (.00043) \\ v2 = 44 \: feet \: per \: second
new velocity coming out of the hose then is
44 ft/sec
4 0
3 years ago
When the velocity of the car is constant the force of friction is?
miv72 [106K]
..... equal to the force from the engine pushing the car forward.

That's why the car is not accelerating.  The horizontal forces on it
are balanced.
6 0
3 years ago
A ski jumper travels down a slope and leaves
Serhud [2]

Question seems to be missing. Found it on google:

a) How long is the ski jumper airborne?

b) Where does the ski jumper land on the incline?

a) 4.15 s

We start by noticing that:

- The horizontal motion of the skier is a uniform motion, with constant velocity

v_x = 28 m/s

and the distance covered along the horizontal direction in a time t is

d_x = v_x t

- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity u_y = 0 and constant acceleration g=9.8 m/s^2 (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is

d_y = \frac{1}{2}gt^2

The time t at which the skier lands is the time at which the skier reaches the incline, whose slope is

\theta = 36^{\circ} below the horizontal

This happens when:

tan \theta = \frac{d_y}{d_x}

Substituting and solving for t, we find:

tan \theta = \frac{\frac{1}{2}gt^2}{v_x t}= \frac{gt}{2v_x}\\t = \frac{2v_x}{g}tan \theta = \frac{2(28)}{9.8} tan 36^{\circ} =4.15 s

b) 143.6 m

Here we want to find the distance covered along the slope of the incline, so we need to find the horizontal and vertical components of the displacement first:

d_x = v_x t = (28)(4.15)=116.2 m

d_y = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(4.15)^2=84.4 m

The distance covered along the slope is just the magnitude of the resultant displacement, so we can use Pythagorean's theorem:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(116.2)^+(84.4)^2}=143.6 m

7 0
3 years ago
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