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3 years ago

Explain, briefly, what the electrical field inside a microwave oven does to heat food.

Physics

1 answer:

yarga [219]3 years ago

5 0

Answer:

Inside the strong metal box, there is a microwave generator called a magnetron. When you start cooking, the magnetron takes electricity from the power outlet and converts it into high-powered, 12cm (4.7 inch) radio waves. Thus the microwaves pass their energy onto the molecules in the food, rapidly heating it up.May 3, 2018

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You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you

vekshin1

Answer:

Approximately $5.9\; {\rm m\cdot s^{-1}}$ (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

the antinode at the end of the spring that is being shaken, and
the antinode between the two ends of this spring.

Overall, the nodes and antinodes on this spring would be:

node at the end that is being held still,
antinode (as mentioned in the question),
node (inferred, not mentioned in the question), and
antinode at the end that is being shaken.

The distance between two adjacent nodes is equal to one-half (that is, $(1/2)$ ) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, $(1/4)$ ) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is $\lambda$ , the length of this spring would be:

$\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda$ .

The question states that the length of this coiled spring is $7.2\; {\rm m}$ . In other words, $(3/4) \, \lambda = 7.2\; {\rm m}$ . The wavelength of this wave would be $(7.2\; {\rm m}) / (3/4) = 9.6\; {\rm m}$ .

The frequency $f$ of this wave is the number of cycles in unit time:

$\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}$ .

Hence, the speed $v$ of this wave would be:

$\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}$ .

3 0

2 years ago

What is a substance

Korolek [52]

Explanation:

A particular kind of matter with uniform properties..

8 0

3 years ago

Read 2 more answers

I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?

Ad libitum [116K]

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

$\frac{1}{2} m {v}^{2} = mgh$

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

$\sin(15) = \frac{h}{d} \\ or \: h = d \sin(15)$

Plugging this into the energy conservation equation and cancelling m, we get

${v}^{2} = 2gd \sin(15)$

Solving for d,

$d = \frac{ {v}^{2} }{2g \sin(15) } = \frac{ {(7.5 \: \frac{m}{s}) }^{2} }{2(9.8 \: \frac{m}{ {s}^{2} })(0.259)} \\ = 11.1 \: m$

3 0

3 years ago

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

3 years ago

A 1.2 KG rubber ball is being thrown in the air if the ball is traveling at 2.0 M/S when it is 3.0 M off the ground what is the

Vitek1552 [10]

Answer:

37.7 J

Hope this helps! (see pictures)

6 0

3 years ago