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Misha Larkins [42]
3 years ago
12

To cook food, a microwave oven converts electrical energy intoenergy A) electromagnetic B) heat C) mechanical D) nuclear

Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
5 0

To cook food, a microwave oven converts electrical energy into electromagnetic energy. (A)  Specifically, it pours high-power RADIO waves into the box.

Water molecules and fat molecules resonate (jiggle) with the radio waves, and the heat is generated IN THE FOOD.

I work with microwave radios.  

-- To communicate high-speed data over a distance of 5 to 10 miles, we use a radio transmitter with a power output of 0.07 watt .

-- To communicate high-speed data over distances up to 80 miles, we use a radio transmitter with a power output of 1 to 2 watts.

-- To heat up a chunk of leftover meatloaf, a microwave oven pumps radio waves into the box with a power output of 1200 to 1500 watts ! !

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7 0
4 years ago
Read 2 more answers
Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
Natali [406]
<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

7 0
3 years ago
How do you think car makers can design cars to limit cell phone distractions?
Dafna11 [192]

I have two (2) brilliant ideas:

1). Inside the metal that the body of the car is made of, and also between the two sheets of glass that the windows are made of, install a thin layer of material that absorbs RF (radio-wave) energy . . . like the material in the glass window of your microwave oven.  Then, no radio waves from the cellular base station can get INTO the car, and no radio waves from your phone can get OUT of the car.  The phone can't make a connection to the cellular network, you can't make or receive calls, and you can't connect to Instagram or Brainly, so you might as well just turn it off and save your battery until next time you're outside your car.

2). Somewhere inside the car, like under the dash or in the glove box, install a teeny tiny radio receiver that can recognize the signals coming OUT of your phone.  Connect it to the car's electrical system so that when it hears signals from phones inside the car, it it shuts down the car's motor so you can't start or drive. The car only works when phones inside the car are either turned off or in Airplane Mode.

My ideas are so brilliant that I really should patent them, or copyright them, or whatever you do so that other people have to pay you to use your idea. But if you want to use them, that's OK.  Just go ahead. I won't mind.

8 0
3 years ago
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

3 0
3 years ago
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