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White raven [17]
2 years ago
9

A dart is thrown at a target that is supported by a wooden backstop. It strikes the backstop with an initial velocity of 350 m/s

[E] The dart comes to rest in 0.0050 s.
a. What is the acceleration of the dart?
b. How far does the dart penetrate into the backstop?​
Physics
1 answer:
11Alexandr11 [23.1K]2 years ago
6 0

Answer:u=350m/sec

v=0m/sec

t=0.005sec

a=v-u/t

=0-350/0.005

= -350/0.005

= -350×1000/5

= -70×1000

= -70000 m/sec^2

Explanation: i think this will help you thankyou

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Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths f
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Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble,  λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t_{min =  ( m + 1/2) × λ/2n

we substitute

t_{min =  ( 0 + 1/2) × 711 /2(1.21)

t_{min = 0.5 × 711/2.42

t_{min = 0.5 × 293.80165

t_{min = 146.9 nm

Therefore,  the minimum wall thickness that will enhance the reflection of light is 146.9 nm

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3 years ago
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NeX [460]

Answer is B. Longitudinal Wave

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Ksju [112]
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The human ear canal is about 2.9 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fu
solniwko [45]

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Explanation:

As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is  

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As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:

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Then the frequency is determined as

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So, the frequency of the human ear canal is 2.92 kHz.

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