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BaLLatris [955]
3 years ago
8

A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for i

t three wet weather on 3 separate days? Also find its Standard Deviation
Physics
1 answer:
lilavasa [31]3 years ago
3 0

Answer:

so the probability will be = 0.062

Standard deviation =  0.8925

Explanation:

The probability of rain = 15% = 15/100= 0.15

and the probability of no rain=q = 1-p= 1-0.15= 0.85

The number of trials = 7

so the probability will be

7C3 * ( 0.15)^3 (0.85)^4= 35* 0.003375 * 0.52200 =0.06166= 0.062

Taking this as binomial as the p and q are constant and also the trials are independent .

For a binomial distribution

Standard deviation = npq= 0.15 *0.85 *7= 0.8925

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Irina finds an unlabeled box of fine needles, and wants to determine how thick they are. A standard ruler will not do the job, a
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Answer:

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by d sin\Theta =1.22\lambda--------eqn 1

The angle through which the first minima is diffracted is given by tan\Theta =\frac{y_1}{D}---------eqn 2

As \Theta is very small so we can write sin\Theta =tan\Theta

So from eqn 1 and eqn 2 we can write

y_1=\frac{1.22\lambda D}{d}--------eqn 3

Here y_1 is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture

It is given that diameter of circular aperture is 14.7 cm so y_1=\frac{14.7}{2}=7.35 \ cm

Now putting all these value in eqn 3

d=\frac{1.22\lambda D}{y_1}

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

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A brick is dropped from a height of 31.9 meters
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A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a
jeka94

Answer:

a)  t_l - t_r = 12.54 us

b)  (t_l - t_r) / T = 0.0157  

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

a. What is the difference in the arrival time of the sound at the left ear and the right ear?

b. What is the ratio of this time difference to the period of the sound wave?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

                                      d_l = sqrt ( 2.6^2 + 0.15^2)

                                      d_l = sqrt ( 6.7825 )

                                      d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

                                      t_l - t_r = (1 / v) * ( d_l - d_r)

Where, v = 343 m/s speed of sound in air:

                                      t_l - t_r = (1 / 343) * ( 2.6043 - 2.6)  

                                      t_l - t_r = ( 0.0043 / 343 )

                                      t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

                                      T = 1 / f

                                      T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

                                      (t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

                                      (t_l - t_r) / T = 0.0157  

3 0
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