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3 years ago

NEED HELP ASAP !!!!!!! anyone gonnna answerrr????

Physics

1 answer:

shepuryov [24]3 years ago

7 0

The 'spot' at the end of the laser beam moves in a circle. The radius of the circle is the distance between the laser and the spot.

The circumference of every circle is (2π) · (radius) .

The speed of the spot is (distance) / (time) .

Speed = (circumference) / (time to turn once around the full circle)

Speed =

(circumference) · (nr of revs) / (second) .

(a). Speed = (2π) (8km) · (9 rev) / sec

Speed = (2π · 8 · 9) km/sec

Speed = 144π km/sec

Speed = 452.4 km/sec

(b). Speed = (2π) (16km) · (9 rev) / sec

Speed = (2π · 16 · 9) km/sec

Speed = 288π km/sec

Speed = 904.8 km/sec

(c). 300,000 km/sec = (2π · distance) · (9 / sec)

300,000 km = (18π · distance)

Distance = 300,000 / 18π km

Distance = 5,305 km

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Answer:

a) 0.159

b) 0.84

Explanation:

The Horizontal component is 2.3 times the vertical component

Let the horizontal electric field component = $E_{h}$

Let the vertical electric field component = $E_{v}$

The formula for light intensity is given by:

$I = \frac{E_{m} ^{2} }{2c \mu}$ ..............................(1)

$E_{m}$ is the resolution of the vertical and horizontal components, $E_{h} and E_{v}$

$E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}$ ..................(2)

Light intensity before the glasses were put on:

$I_{1} = \frac{E_{m} ^{2} }{2c \mu_{1} }$ .............................(3)

Put equation (2) into equation (3)

$I_{1} = \frac{E_{h} ^{2} + E_{v} ^{2}}{2c \mu_{1} }$ .............................(4)

After the glasses were put on the horizontal component vanishes, i.e. $E_{h} = 0$

$I_{2} = \frac{ E_{v} ^{2}}{2c \mu_{2} }$ ...................................(5)

Divide equation (5) by equation (4)

$\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$ ...............................(6)

But $E_{h} = 2.3E_{v}$ ......................(7)

Insert equation (7) into (6)

$\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{(2.3E_{v})^{2} + E_{v} ^{2} } \\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{5.29E_{v}^{2} + E_{v} ^{2} }\\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29} \\$

$\frac{I_{2} }{I_{1} }= 0.159$

b) When the sunbather lies on his side, the vertical component vanishes, i.e $E_{v} = 0$

$\frac{I_{2} }{I_{1} } = \frac{E_{h} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$

$\frac{I_{2} }{I_{1} } = \frac{(2.3E_{v} )^{2} }{E_{v} ^{2} +(2.3E_{v} )^{2}}$

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{E_{v} ^{2} +5.29E_{v}^{2} }$

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } = \frac{5.29}{6.29} \\\frac{I_{2} }{I_{1} } = 0.84$

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3 years ago

Lesechka [4]

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Explanation:

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3 years ago

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Evgen [1.6K]

The frequencies are missing in the question. The three successive resonance frequencies within the volume are $f_1, f_2 \text{ and}\ f_3$ .

Solution :

Let the volume be : v

The frequency for one end open and one end closed is given by :

So, $f_n = \frac{(2n-1)v}{4L}$

Therefore,

$f_1 = \frac{v}{4L}$ , $f_2 = \frac{3v}{4L}$ , $f_3 = \frac{5v}{4L}$

So, $f_2 - 3f_1$ and $f_3=\frac{5}{3}f_2$

Therefore, the ratio of $\frac{f_3}{f_2}=\frac{5}{3}$ which is not a whole number.

Now the frequency for open volume and closed at both end

$f_n=\frac{nv}{4L}$

So, $f_1=\frac{v}{4L}$ , $f_2 = 2f_1$ , $f_2=3f_1$

From above formulae we can see that ratio of is not a whole number that is a identification for the frequency of the volume at one end open and one end closed.

Also ratio of the consecutive frequency of the volume at open from both side and closed from both side is always a whole number.

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3 years ago