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Kryger [21]
3 years ago
5

A metallic sheet has a large number of slits, 5.0 mm wide and 20 cm apart, and is used as a diffraction grating for microwaves.

A wide parallel beam of microwaves is incident normally on the sheet. What is the smallest microwave frequency for which only the central maximum occurs?
Physics
1 answer:
san4es73 [151]3 years ago
7 0

To solve this problem it is necessary to apply the concepts related to the principle of superposition and constructive interference, that is to say everything that refers to an overlap of two or more equal frequency waves, which when interfering create a new pattern of waves of greater intensity (amplitude) whose cusp is the antinode.

Mathematically its definition can be given as:

d sin\theta =m\lambda

Where

d = Width of the slit

\theta = Angle between the beam and the source

m = Order (any integer) which represent the number of repetition of the spectrum, at this case 1 (maximum respect the wavelength)

Since the point of the theta angle for which the diffraction becomes maximum will be when it is worth one then we have to:

\lambda = d

\lambda = 20cm = 20*10^{-2}m

Applying the given relation of frequency, speed and wavelength then we will have that the frequency would be:

f = \frac{v}{\lambda}

Here the velocity is equal to the speed of light and the wavelength to the value previously found.

f = \frac{3*10^8}{0.2}

f = 1.5Ghz

Therefore the smallest microwave frequency for which only the central maximum occurs is 1.5Ghz

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Can y’all please help me with this 3 part question?
klio [65]

Answer:

Vf = 210 [m/s]

Av = 105 [m/s]

y = 2205 [m]

Explanation:

To solve this problem we must use the following formula of kinematics.

v_{f} =v_{o} +g*t

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (released from the rest)

g = gravity acceleration = 10 [m/s²]

t = time = 21 [s]

Vf = 0 + (10*21)

Vf = 210 [m/s]

Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)

The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)

Av = (210 + 0)/2

Av = 105 [m/s]

To calculate the distance we must use the following equation of kinematics

v_{f} ^{2} =v_{o} ^{2} +2*g*y\\\\(210)^{2} = 0 + (2*10*y)

44100 = 20*y

y = 2205 [m]

5 0
3 years ago
Which of the following statements about diodes and triodes is correct
nata0808 [166]
There are no correct statements on your list.
4 0
3 years ago
A 125 W motor accelerates a block along a level, frictionless surface at an average speed of 5.0
Lelu [443]

Answer:

25N

Explanation:

Given parameters:

Power of the motor = 125W

Average speed  = 5m/s

Unknown:

Force supplied to the motor = ?

Solution:

 Work done is the force applied to move a body through a particular distance;

         Work done  = Force x distance

Also,

        Work done  = Power x time

 

So;

             Force x distance = Power  x time

  since force is the unknown;

            Force  = \frac{Power x time}{distance}

         Speed  = \frac{distance }{time}

             Force  = \frac{Power}{speed}

Now solve;

                Force  = \frac{125}{5}    = 25N

3 0
3 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 34.3 m/s^2 . The acc
Eddi Din [679]

Answer:

The maximum height reached by the rocket is 1.94 × 10³ m.

Explanation:

The height of the rocket can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t²    (when the rocket is accelerated upward).

y = y0 +  v0 · t + 1/2 · g · t² (after the rocket runs out of fuel).

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to engines of the rocket.

g = acceleration due to gravity.

In the same way, the velocity of the rocket can be calculated as follows:

v = v0 + a · t  (when the rocket has fuel)

v = v0 + g · t   (when the rocket runs out of fuel)

Where "v" is the velocity at time "t"

First, let´s find the height reached until the rocket runs out of fuel.

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · t + 1/2 · 34.3 m/s² · (5.00 s)²

y = 429 m

And now, let´s find the velocity reached in that time of upward acceleration:

v = v0 + a · t

v = 0 m/s + 34.3 m/s² · 5.00 s

v = 172 m/s

When the rocket runs out of fuel, it is accelerated downward due to gravity. But, since the rocket has initially an upward velocity (172 m/s), it will not fall immediately and will continue to go up until the velocity becomes 0. In that instant, the rocket is at its maximum height and thereafter it will start to fall with negative velocity.

Then, using the equation for velocity, we can calculate the time it takes the rocket to reach its maximum height:

v = v0 + g · t

0 = 172 m/s - 9.80 m/s² · t

-172 m/s / -9.80 m/s² = t

t = 17.6 s

With this time, we can now calcualte the maximum height. Notice that the initial velocity and height are the ones reached during the upward acceleration phase:

y = y0 +  v0 · t + 1/2 · g · t²

ymax = 429 m + 172 m/s · 17.6 s - 1/2 · 9.80 m/s² · (17.6 s)²

ymax = 1.94 × 10³ m

4 0
3 years ago
A student that jumps a vertical height of 50 cm during the hang time activity.
muminat

The hang time of the student is 0.64 seconds, and he must leave the ground with a speed of 3.13 m/s

Why?

To solve the problem, we must consider the vertical height reached by the student as max height.

We can use the following equations to solve the problem:

<u>Initial speed calculations:</u>

v_{f}^{2}=v_{o}^{2}+2*a*d

At max height, the speed tends to zero.

So, calculating, we have:

<u>v_{f}^{2}=v_{o}^{2}+2*a*d\\\\0=v_{o}^{2}+2*(-9.81\frac{m}{s^{2}})*0.5m\\\\v_{o}^{2}=9.81\frac{m^{2} }{s^{2}}\\\\v_{o}=\sqrt{9.81\frac{m^{2} }{s^{2}}}=3.13\frac{m}{s}</u>

<u>Hang time calculations:</u>

We must remember that the total hang time is equal to the time going up plus the time going down, and both of them are equal,so, calculating the time going down, we have have:

y-yo=vo.t+\frac{1}{2}*9.81\frac{m}{s^{2}}*t^{2} \\\\0.5m-0=0*t+4.95*t^{2}\\\\t^{2}=\frac{0.5}{4.95}\\\\t=\sqrt{0.101}=0.318s

Then, for the total hang time, we have:

TotalHangTime=2*0.318seconds=0.64seconds

Have a nice day!

3 0
4 years ago
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