All categories

3 years ago

What are at least three types of energy involving a microwave

1 answer:

8 0

I'm pretty sure what you are trying to ask for is radiative energy, light energy, and electronic energy.
Radiative since the microwave is releasing radiation,
Light since there is light inside the microwave,
Electronic since it is plugged in and uses electricity.
You can also use sound, but I don't think every microwave makes sound.

You might be interested in

ASHA 777 [7]

Answer:

Explanation:

6+10=16

the box will go forward but it will be a little harder.

6 0

3 years ago

A wire of resistivity ρ must be replaced in a circuit by a wire of the same material but four times as long. If, however, the to

Elanso [62]

Answer:

E. two times the original diameter

Explanation:

Resistance of a wire is:

R = ρ L/A

where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area.

For a round wire with diameter d:

R = ρ L / (¼ π d²)

The two wires must have the same resistance, so:

ρ₁ L₁ / (¼ π d₁²) = ρ₂ L₂ / (¼ π d₂²)

The wires are made of the same material, so ρ₁ = ρ₂:

L₁ / (¼ π d₁²) = L₂ / (¼ π d₂²)

The new length is four times the old, so 4 L₁ = L₂:

L₁ / (¼ π d₁²) = 4 L₁ / (¼ π d₂²)

1 / (¼ π d₁²) = 4 / (¼ π d₂²)

Solving:

1 / (d₁²) = 4 / (d₂²)

(d₂²) / (d₁²) = 4

(d₂ / d₁)² = 4

d₂ / d₁ = 2

So the new wire must have a diameter twice as large as the old wire.

3 0

3 years ago

Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not

musickatia [10]

Answer:

Part A: $t = v_0/a_0$

Part B: $t = v_0/a_0$

Part C: $v_0^2/a_0$

Explanation:

Part A:

We will use the following kinematics equation:

$v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}$

Part B:

We will use the same kinematics equation:

$v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}$

Part C:

The total time takes is 2t.

So the train moves a distance of

$x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}$

And the car moves a distance in Part A and in Part B:

$d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}$