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4 years ago

7

During an observation, it was noticed that light diffracts as it passes through small slits in a barrier. What does this evidenc

e reveal about light? (2 points)

1 answer:

Lynna [10]4 years ago

8 0

Answer:

It reveals that light is a wave

Explanation:

Diffraction is the property of a wave in which there is a bending of the wave about the corners of an obstacle or aperture into the geometrical shadow of the obstacle or aperture.

This simply implies that a wave bends or spreads out when it passes through openings. Since the light diffracts through small slits and diffraction has been shown to occur in water waves and sound waves, this property of diffraction can only be characteristic of a wave and thus, this evidence reveals that light is a wave.

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Answer:

v = 5.75 x 10⁶ m/s

Explanation:

The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by

r = $\frac{mv}{qB}$ --------------(i)

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m = mass of the particle

Making v subject of the formula in equation (i) above gives

v = $\frac{qBr}{m}$ -------------------(ii)

Given;

r = 20cm = 0.2m

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Substitute the values of m, q, B and r into equation (ii) above to get;

v = $\frac{1.6 * 10^{-19} * 0.3 * 0.2} {1.67*10^{-27} }$

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3 years ago

A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s2 . It runs out of fuel at the end of 4.00s

Murrr4er [49]

Answer:

a) The velocity and position of the rocket at the end of 4 seconds are 117.6 meters per second and 235.2 meters, respectively, b) The maximum height reached by the rocket is 940.296 meters, c) The rocket crashes on the ground at a velocity of -96.030 meters per second.

Explanation:

The complete statement is:

A rocket moves straight upward , starting from rest with an acceleration of 29.4 m/s2. it runs out of fuel at the end of 4.00 s and countinues to coast upward , reaching a maximum height before falling back to earth . a) find the rocket's velocity and position at the end of 4.00 s . b) Find the maximum height the rocket reaches. c) find the velocity on the instant before the rocket crashes on the ground.

a) The rocket accelerates uniformly, whose equations of motion are:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot a \cdot t^{2}$

$v = v_{o} + a\cdot t$

Where:

$y$ - Final position, measured in meters.

$y_{o}$ - Initial position, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

$t$ - Time, measured in second.

$a$ - Acceleration, measured in meters per square second.

$v$ - Final velocity, measured in meters per second.

If we know that $y_{o} = 0\,m$ , $v_{o} = 0\,\frac{m}{s}$ , $t = 4\,s$ and $a = 29.4\,\frac{m}{s^{2}}$ , the velocity and position of the rocket are, respectively:

$y = 0\,m+\left(0\,\frac{m}{s} \right)\cdot (4\,s)+\frac{1}{2}\cdot \left(29.4\,\frac{m}{s^{2}} \right) \cdot (4\,s)^{2}$

$y = 235.2\,m$

$v = 0\,\frac{m}{s} + \left(29.4\,\frac{m}{s^{2}}\right)\cdot (4\,s)$

$v = 117.6\,\frac{m}{s}$

The velocity and position of the rocket at the end of 4 seconds are 117.6 meters per second and 235.2 meters, respectively.

b) Now, the rocket experiments a free-fall motion. The maximum height of the rocket is obtained by equalizing the equation of velocity to zero and evaluating the equation of position later. That is:

$y = 235.2\,m+\left(117.6\,\frac{m}{s} \right)\cdot t+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}$

$v = 117.6\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right)\cdot t$

Then,

$0 = 117.6 -9.807\cdot t$

$t = 11.991\,s$

The maximum height reached by the rocket is:

$y = 235.2+117.6\cdot (11.991)+\frac{1}{2}\cdot (-9.807)\cdot (11.991)^{2}$

$y = 940.296\,m$

The maximum height reached by the rocket is 940.296 meters.

c) The rocket experiments a free-fall motion and is accelerated by gravity until collision happens. The equations of motion below are presented:

$0\,m = 940.296\,m+\left(0\,\frac{m}{s} \right)\cdot t+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}$

$v = 0\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right)\cdot t$

From the equation of position we get the instant when rocket hits the ground, whose roots are found by Quadratic Formula:

$t_{1} \approx 9.792\,s$ and $t_{2} \approx -9.792\,s$

Only the first root is physically reasonable.

By the second equation we calculate the final velocity:

$v = 0-9.807\cdot (9.792)$

$v = -96.030\,\frac{m}{s}$

The rocket crashes on the ground at a velocity of -96.030 meters per second.

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4 years ago