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3 years ago

13

A microwave works by focusing microwave light on the food inside of it. Which type of energy transformation takes place in a mic

rowave? A. Heat energy is transformed into electrical energy. B. Light energy is transformed into electrical energy. C. Electrical energy is transformed into light energy. D.

2 answers:

inysia [295]3 years ago

7 0

First, electrical energy is transformed into electromagnetic energy ... the microwave radio waves (which you called 'light').

After that, the microwave radio waves carry electromagnetic energy into the meatloaf, where it's transformed into heat energy.

Sveta_85 [38]3 years ago

4 0

Electrical energy is transformed into light energy.

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These spectra are from the same element. Which is an emission spectrum, which an absorption spectrum?

RSB [31]

Answer:

D. Top is emission; bottom absorption.

Explanation:

Emission and spectrum of elements are due to the element absorbing or emitting wavelength of e-m energy. Elementary particles of elements can absorb energy from a ground state to enter an excited state, creating an absorption spectrum, or they can lose energy and fall back to a lower energy state, creating an emission spectrum. A simple rule to differentiate between an emission and an absorption spectrum is that: "all absorbed wavelength is emitted, but not all emitted wavelength is absorbed."

From the image, the lines indicates wavelengths. We can see that all of the wavelengths of the bottom absorption spectrum coincides with some of the wavelength of the upper emission wavelengths.

3 0

3 years ago

Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t

Readme [11.4K]

Answer: $3.66(10)^{33}kg$

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity $\omega$ of the planet P1 with a period $T=750years=2.36(10)^{10}s$ :

$\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}$ (1)

Where:

$V_{1}=40.2km/s=40200m/s$ is the velocity of planet P1

$R$ is the radius of the orbit of planet P1

Finding $R$ :

$R=\frac{V_{1}}{2\pi}T$ (2)

$R=\frac{40200m/s}{2\pi}2.36(10)^{10}s$ (3)

$R=1.5132(10)^{14}m$ (4)

On the other hand, we know the gravitational force $F$ between the star S with mass $M$ and the planet P1 with mass $m$ is:

$F=G\frac{Mm}{R^{2}}$ (5)

Where $G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

In addition, the centripetal force $F_{c}$ exerted on the planet is:

$F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}$ (6)

Assuming this system is in equilibrium:

$F=F_{c}$ (7)

Substituting (5) and (6) in (7):

$G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}$ (8)

Finding $M$ :

$M=\frac{V^{2}R}{G}$ (9)

$M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}$ (10)

Finally:

$M=3.66(10)^{33}kg$ (11) This is the mass of the star S

4 0

3 years ago

Four objects are situated along the y axis as follows: a 1.93-kg object is at +2.91 m, a 2.95-kg object is at +2.43 m, a 2.41-kg

Anna [14]

Answer:

0.958 m

Explanation:

So the total mass of the system is

M = 1.93 + 2.95 + 2.41 + 3.99 = 11.28 kg

let y be the distance from the center of mass to the origin. With the reference to the origin then we have the following equation

$My = m_1y_1 + m_2y_2 +m_3y_3 + m_4y_4$

$11.28y = 1.93*2.91 + 2.95*2.43 + 2.41*0 + 3.99*(-0.496) = 10.806$

$y = \frac{10.806}{11.28} = 0.958 m$

So the center of mass is 0.958 m from the origin

3 0

3 years ago

A 58 kg boy and a 38 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface. If the acceleratio

Arada [10]

H rerrr hrdzjrtfhhtthfjytr

6 0

3 years ago

6. A suitcase weighing 120 N sits on a counter 1.8 m high

Liono4ka [1.6K]

Answer:

<em>216 J</em>

Explanation:

h = 1.8

a = 9.8

m = 12.2

<em>GPE</em> = <em>HAM</em> = 216

6 0

2 years ago