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Dmitry_Shevchenko [17]
3 years ago
9

The speed of a light wave in a certain transparent material is 0.589 times its speed in vacuum, which is 3.00 x108 m/s. When yel

low light with a frequency of 5.25 x104 Hz passes through this material, what is its wavelength in nanometers? Number 囗 n m
Physics
1 answer:
irakobra [83]3 years ago
8 0

Answer:

The wavelength of the yellow light in given transparent material is <em><u>337 nm</u></em>

Explanation:

Speed of the light in vacuum , c=3.00\times 10^{8}\frac{m}{s}

Therefore speed of light in the given material is v=0.589\times c

=>v=0.589\times 3.00\times 10^{8}\frac{m}{s}=1.767\times 10^{8}\frac{m}{s}

frequency of yellow light , f=5.25\times 10^{14}Hz

Wavelength in the given transparent material , \lambda =\frac{v}{f}=\frac{1.767\times 10^{8}}{5.25\times 10^{14}}m

=> \lambda =3.366\times 10^{-7}m=336.6nm

=>\lambda \simeq 337nm

Thus the wavelength of the yellow light in given transparent material is <em><u>337 nm</u></em>

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<h2>Answer:</h2>

-310J

<h2>Explanation:</h2>

The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e

ΔE = Q + W       ----------------------(i)

If heat is released by the system, Q is negative. Else it is positive.

If work is done on the system, W is positive. Else it is negative.

<em>In this case, the system is the balloon and;</em>

Q = -0.659kJ = -695J    [Q is negative because heat is removed from the system(balloon)]

W = +385J  [W is positive because work is done on the system (balloon)]

<em>Substitute these values into equation (i) as follows;</em>

ΔE = -695 + 385

ΔE = -310J

Therefore, the change in internal energy is -310J

<em>PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.</em>

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5 0
3 years ago
provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small
gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

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Also, the angular frequency of physical pendulum is;

ω = √(mgL / I ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and I  is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

I = \frac{1}{3}mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/I) OR T = 2π√(I/mgL)

so we can use I = \frac{1}{3}mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L =  \frac{1}{2}D.

now, substituting these equations, the period becomes;

T = 2π/√(I/mgL) OR T = 2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } } OR T = 2π√(2D/3g )  ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω_{simple = 2π/T_{simple OR  ω_{simple = √(g/D) OR  ω_{simple = 2π√( D/g )  

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × T_{simple

we substitute in value of T_{simple

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

 

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Vitek1552 [10]
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