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Iteru [2.4K]
3 years ago
5

What is the wavelength in nm of a light whose first order bright band forms a diffraction angle of 30 degrees, and the diffracti

on grating has 700 lines per mm? ​
Physics
1 answer:
Artist 52 [7]3 years ago
8 0

Answer:

The wavelength is 3500 nm.

Explanation:

d= \frac{1}{700 lines per mm} = 0.007mm = 7000 nm

n= 1

θ= 30°

λ= unknown

Solution:

d sinθ = nλ

λ = \frac{7000 nm sin 30}{1}

λ = 3500 nm

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Which steps can be taken to translate the phrase “the height of a tree is increased by seven inches” into an algebraic expressio
ololo11 [35]

Answer:

Explanation:

height of tree = x

seven inches= 7

increaced by=+

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2 years ago
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The maximum pressure variations the human ear can withstand are about how much above and below atmospheric pressure?
Crank

The maximum pressure variations the human ear can withstand  above and below atmospheric pressure is around 30 pa. the normal atmospheric pressure is around 101325 pa. hence the variation in the maximum pressure for human ear is very small as compared to the atmospheric pressure. if the ear is exposed to a pressure greater than this , it can cause permanent damage to the ear.

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Which of the following is an example of the media's influence on government? (3 points)
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3 years ago
Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a
Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

<em>consider the attached diagram below</em>

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }

Sum of the potential at point p is

V = Va + Vb

that is

V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }

V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }

V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]

the expression below can be written as the equivalent

\frac{1}{(a^{2} + r^{2} )^{1/2} }  = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }

likewise,

\frac{1}{(b^{2} + r^{2} )^{1/2} }  = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }

hence,

V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]

by reciprocal rule

1/a² = a⁻²

V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]

by binomial expansion of fractional powers

where (1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...

if we expand the expression we have the equivalent as shown

{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )

also,

{(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )

the above equation becomes

V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]

V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]

V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]

V = \frac{q}{4\pi e0 r} * \frac{1}{2r^{2} } *(b^{2} -a^{2} )

V = \frac{q}{8\pi e0 r^{3} } * (b^{2} -a^{2} )

Answer

V = \frac{q (b^{2} -a^{2} )}{8\pi e0 r^{3} }

OR

V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }

8 0
3 years ago
What is needed to set a body in motion
kumpel [21]

Answer:

Newtons law

Explanation:

According to this law, a body at rest tends to stay at rest, and a body in motion tends to stay in motion, unless acted on by a net external force.

3 0
3 years ago
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