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inessss [21]
3 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]3 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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IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s

Time taken by the ball to reach the highest point is 0.14 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft

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8 0
4 years ago
What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?
goldfiish [28.3K]

Answer:

1.05 J.

Explanation:

Kinetic Energy: This is the energy possessed by a body due to its motion. The S.I unit of kinetic energy is Joules (J). The formula of kinetic energy is given as

Ek = 1/2mv²................. Equation 1

Where Ek = kinetic energy, m = mass of the uniform rod, v = liner velocity of the rod.

But,

v = αr .......................... Equation 2

Where α = angular velocity of the rod, r = radius of the circle.

Given: α = 3.6 red/s, r = 120/2 = 60 cm = 0.6 m.

Substitute into equation 2

v = 3.6(0.6)

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Also given: m = 450 g = 0.45 kg.

Substitute into equation 1

Ek = 1/2(0.45)(2.16²)

Ek = 1.05 J.

4 0
3 years ago
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C the runners feet pushing against the ground describes the acceleration toward the finish line
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Pani-rosa [81]

Answer:

Power is the rate which work is done.

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