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inessss [21]
2 years ago
15

The microwaves in a certain microwave oven have a wavelength of 12.2 cm. How wide must this oven be so that it will contain five

antinodal planes of the electric field along its width in the standing wave pattern?
Suppose a manufacturing error occurred and the oven was made 4.0 cm longer than specified in part (a). In this case, what would have to be the frequency of the microwaves for there still to be five antinodal planes of the electric field along the width of the oven?
Physics
1 answer:
leva [86]2 years ago
8 0

Answer:

a

 l = 0.305 \  m

b

  f = 3.0*10^{11} \  Hz

Explanation:

From the question we are told that

  The  wavelength is  \lambda  =  12.2 \  cm  = 0.122 \  m

  The  number of antinodal planes of the electric field considered is n  =  5

The  width is mathematically represented as

       l  =  \frac{ n \lambda}{2}

       l = \frac{5 * 0.122 }{ 2}

      l = 0.305 \  m

Generally the  frequency the errors was made is  mathematically represented as

   f =  \frac{c}{\lamda_k}

Here c is the speed of light with value  c =  3.0*10^{8} \  m/s

     \lambda_k is the wavelength of the microwave has to be in order for there still to be five antinodal planes of the electric field along the width of the oven, which is mathematically represented as

     \lambda_k  =  \frac{ \lambda *  \frac{0.04}{2} }{n/2}

      \lambda_k  =  \frac{0.122*0.02}{5/2}

So

   f =  \frac{3.0*10^{8}}{0.000976}

    f = 3.0*10^{11} \  Hz

   

       

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Salmon often jump waterfalls to reach their
natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

v_x = u cos \theta

where u is the initial speed and \theta=37.7^{\circ}. The horizontal distance travelled by the salmon is

d=v_x t = (ucos \theta)t

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

t=\frac{d}{v_x}=\frac{d}{u cos \theta} (1)

Along the vertical direction, the equation of motion is

y=h+u_y t -\frac{1}{2}gt^2

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

u_y = u sin \theta is the vertical component of the initial velocity of the salmon

g=9.81 m/s^2 is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}

and we can solve this formula for u, the initial speed of the salmon:

y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s

5 0
3 years ago
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Aloiza [94]
False. because the object cannot freefall due to resistance, so it cannot develop full acceleration.
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Does coefficient of thermal expansion vary with temperature?
N76 [4]

Answer:

yes

Explanation:

8 0
2 years ago
What is the speed of light in a vacuume?
-Dominant- [34]

The speed of light in vacuum is exactly

299,792,458 meters per second.

That's so exact that it's the official scientific definition of a "meter".

The number doesn't t change even if the flashlight or other light source is moving.

5 0
2 years ago
In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help
Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

4 0
3 years ago
Read 2 more answers
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