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DanielleElmas [232]
3 years ago
15

A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 37.6 vibrations

in 27.9 s. Also, a given maximum travels 450 cm along the rope in 11.3 s. What is the wavelength? Answer in units of cm.
Physics
1 answer:
tangare [24]3 years ago
8 0

Answer:

\lambda = 25.79\ cm

Explanation:

given,

Wave vibrates = 37.6

time = 27.9 s

maximum distance travel = 450 cm

time = 11.3 s

wavelength = ?

frequency of wave

f=\dfrac{37.6}{27.9}

f = 1.35 Hz

Speed of wave

v = \dfrac{450}{11.3}

v = 39.82 cm/s

wavelength of wave

v = fλ

\lambda =\dfrac{v}{f}

\lambda =\dfrac{34.82}{1.35}

\lambda = 25.79\ cm

Hence, wavelength of the wave is equal to 25.79 cm.

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Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

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Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

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u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

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Vector A has magnitude of 15.0 m/s and is 75° counter-clockwise up from the x-axis. What are the x- and y-components of the vect
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Now x-component will be 15 cos75°=3.8822 ( as it makes an angle of 75° with x-axis )

y- component will be 15 sin 75°=14.488

For verification the resultant of x and y component should be equal to 15

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This normal ray is a ray that is perpendicular to the surface.

According to the question, if the beam of light is reflected off the surface and its angle of incidence is 30°, its angle of reflection will also be 30° i.e i=r = 30°

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