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slava [35]
2 years ago
13

The speed of a moving bullet can be deter-

Physics
1 answer:
Bad White [126]2 years ago
7 0

Answer:

<em>v = 381 m/s</em>

Explanation:

<u>Linear Speed</u>

The linear speed of the bullet is calculated by the formula:

\displaystyle v=\frac{x}{t}

Where:

x = Distance traveled

t = Time needed to travel x

We are given the distance the bullet travels x=61 cm = 0.61 m. We need to determine the time the bullet took to make the holes between the two disks.

The formula for the angular speed of a rotating object is:

\displaystyle \omega=\frac{\theta}{t}

Where θ is the angular displacement and t is the time. Solving for t:

\displaystyle t=\frac{\theta}{\omega}

The angular displacement is θ=14°. Converting to radians:

\theta=14*\pi/180=0.2443\ rad

The angular speed is w=1436 rev/min. Converting to rad/s:

\omega = 1436*2\pi/60=150.3776\ rad/s

Thus the time is:

\displaystyle t=\frac{0.2443\ rad}{150.3776\ rad/s}

t = 0.0016 s

Thus the speed of the bullet is:

\displaystyle v=\frac{0.61}{0.0016}

v = 381 m/s

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A 500 lines per mm diffraction grating is illuminated by light of wavelength 580 nm . what is the maximum diffraction order seen
vampirchik [111]

The maximum diffraction order seen is 3.

<h3>What is the maximum diffraction order seen?</h3>

We know that the maximum angle of diffraction Q_m of the furthest bright fringe < Q = 90 degrees.

Here we need to compute the nth bright fringe for which is approximated to 90 degrees.

The angle of nth bright fringe is given by;

sin(Q_m) = n(λ)N

Approximating Q_m ≈ 90 degrees.

sin (90) = nλN

n = sin (90) / (λN)

n = 1 / ((580 x 10⁻⁶)500)

n = 3.5 orders

Since, we knew that Q_m < 90 degrees, we will choose n = 3 as the maximum number of orders.

Thus, the maximum diffraction order seen is 3.

Learn more about maximum diffraction here: brainly.com/question/14703089

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2 years ago
What happens when velocity and acceleration are at right angles to each other
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What effect do sound waves have on air particles
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Answer:

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Explanation:

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3 years ago
A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down
rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

V = \sqrt{\frac{T}{\rho}}

Where,

T = Tension

\rho = Linear density

Our data are given by

Tension , T = 70 N

Linear density , \rho = 0.7 kg/m

Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

V = \sqrt{\frac{T}{\rho}}

V = \sqrt{\frac{70}{0.7}

V = 10m/s

Speed can also be expressed as

V = \lambda f

Re-arrange to find \lambda

\lambda = \frac{V}{f}

Where,

f = Frequency,

Which is also described in function of the Period as,

f = \frac{1}{T}

f = \frac{1}{0.35}

f = 2.86 Hz

Therefore replacing to find \lambda

\lambda = \frac{10}{2.86}

\lambda = 3.49m

Therefore the wavelength of the waves created in the string is 3.49m

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3 years ago
Which feature of the sun appears in cycles of about 11 years?
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