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3 years ago

How is a light wave likely to react when hitting a shiny opaque object?

Physics

1 answer:

shepuryov [24]3 years ago

6 0

Answer:

Explanation:

The light wave will be reflected

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Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage

agasfer [191]

$v = 2.45×10^3\:\text{m/s}$

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

$F = \dfrac{d{p}}{d{t}}$ (1)

Assuming that the velocity remains constant then

$F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}$

Solving for $v,$ we get

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)$

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

$F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}$

The exhaust rate dm/dt is

$\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}$

$\;\;\;\;\;= 1.36×10^4\:\text{kg/s}$

Therefore, the velocity at which the exhaust gases exit the engines is

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}$

$\;\;\;= 2.45×10^3\:\text{m/s}$

6 0

2 years ago

What is the wavelength that corresponds to a frequency of 6.00x1014 hz?

kipiarov [429]

The basic relationship between wavelength $\lambda$ , frequency f and speed c of an electromagnetic wave is
$\lambda= \frac{c}{f}$
where c is the speed of light. Substituting numbers, we find:
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{6.0 \cdot 10^{14} Hz}=5\cdot 10^{-7} m$

4 0

3 years ago

Explain how carbon dioxide, in its standard state, would compare to the solubilities shown on the graph.

Sladkaya [172]

The answer is C. It would look similar to the graph for KNO3

8 0

2 years ago

Read 2 more answers

An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

Zigmanuir [339]

Answer:

(A) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )$

$\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )$

$\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

7 0

3 years ago

Nuclear energy is the energy stored in the of an atom. t/f

Shtirlitz [24]

In nuclear fission, atoms are split to release the energy. A nuclear reactor, or power plant, is a series of machines that can control nuclear fission to produce electricity. The fuel that nuclear reactors use to produce nuclear fission is pellets of the element uranium. So it would be false.

3 0

2 years ago

Read 2 more answers