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timama [110]
2 years ago
9

A microwave oven operates at 3.00 ghz . what is the wavelength of the radiation produced by this appliance?

Physics
1 answer:
Nadya [2.5K]2 years ago
3 0
Thank you for posting your question here at brainly. Below is the solution:

Speed of light is 300,000,000 meters per second or 3E8 m/s 

<span>Wavelength = Speed of light divided by the frequency = c / f </span>

<span>3 GHz = 3 GigaHertz = 3E9 Hertz = 3,000,000,000 Hz </span>

<span>L = (300,000,000 m/s) / (3,000,000,000 Hz) = 0.1 meter = 10 centimeters </span>
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A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0
2 years ago
Consider two waves defined by the wave functions y1(x,t)=0.50msin(2π3.00mx+2π4.00st) and y2(x,t)=0.50msin(2π6.00mx−2π4.00st). Wh
guapka [62]

Answer:

They two waves has the same amplitude and frequency but different wavelengths.

Explanation: comparing the wave equation above with the general wave equation

y(x,t) = Asin(2Πft + 2Πx/¶)

Let ¶ be the wavelength

A is the amplitude

f is the frequency

t is the time

They two waves has the same amplitude and frequency but different wavelengths.

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2 years ago
block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th
Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

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Fnet = 0.5 × -30

Fnet = -15N

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The net force is directed to the negative direction of the x -axis

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Arlecino [84]
F=ma
f?
m=1300kg
a=1.07m\s squared
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2 years ago
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