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valentinak56 [21]
2 years ago
9

As a pan is sitting on a stove it begins to heat up. Eventually it may get hot enough that it glows red, meaning it is radiating

electromagnetic waves within the visible portion of the spectrum. What does this situation show us about electromagnetic radiation?
Electromagnetic radiation can only be emitted in the form of visible light.
The hotter an object is, the faster the electromagnetic radiation will travel.
The electromagnetic radiation emitted by an object is not affected by its temperature.
The hotter an object is, the greater the energy of the electromagnetic radiation in emits.
Physics
1 answer:
saul85 [17]2 years ago
4 0
Answer:
Your answer is B.!

When things in general get hotter, the atoms inside move faster and vise versa with cold.

(for example water); if you put water in a pot on a stove to boil, its gets so hot to where the atoms are moving really fast to get it to look like that.

Please mark brainiest! if this answer is wrong, (which I'm pretty sure it isn't) then pick D. :D
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The y-position of a damped oscillator as a function of time is shown in the figure.
NISA [10]

(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

<h3>Damping coefficient</h3>

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

<h3>at time, t = 0, y = 3.5</h3>

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

<h3>at time, t = 1 cm, y = - 3cm</h3>

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) =  e^(-bx)

ln[-0.857 / cos(ω)] = -bx  

ln[-0.857 / cos(ω)] / b = - x  ---- (1)

<h3>at time, t = 2 cm, y = - 2cm</h3>

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx  

ln[-0.57 / cos(2ω)] /2b = - x  ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) =  cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω  - cosω - 6 = 0

let cosω  = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω  = arc cos(-0.92)

ω  = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b  ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

Learn more about damping coefficient here: brainly.com/question/14058210

#SPJ1

4 0
2 years ago
RP 1 - Specific Heat Capacity GCSE Exam Questions (B.S.G)
Stels [109]

When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.

A sample of 0.50 kg of water boils (reaches 100 °C). After a while, its temperature decreases by 22 °C.

We can calculate the energy transferred to the surroundings from the water in the form of heat (Q) using the following expression.

Q = c \times m \times \Delta T = \frac{4200J}{kg.\° C}  \times 0.50kg \times (-22\° C) \times \frac{1kJ}{1000J} = -46.2 kJ

where,

  • c: specific heat capacity of water
  • m: mass of water
  • ΔT: change in the temperature

When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.

Learn more: brainly.com/question/16104165

8 0
3 years ago
Classify each change (which can be manipulated within the green box) according to its effect on the wavelength.
Nataly_w [17]

Answer:

Explanation:

The classification will be made into 3 categories, which are

Ones that shortens wavelengths

Ones that lengthens wavelengths

Ones that has no effect on wavelengths

Shortens wavelengths -> Increase frequency

Lengthens wavelengths -> Decrease frequency

No effect -> Increase amplitude, decrease amplitude, increase damping, decrease damping.

7 0
3 years ago
A laser beam is incident on two slits with a separation of 0.230 mm, and a screen is placed 4.75 m from the slits. If the bright
Mariulka [41]

Answer:

7.55\times 10^{-7} m

Explanation:

We are given that

d=0.23 mm=0.23\times 10^{-3} m

1mm=10^{-3} m

Screen is placed  from the slits at distance ,L=4.75 m

The bright interference fringes on the screen are separated  by 1.56 cm.

\Delta y=1.56 cm=1.56\times 10^{-2} m

1 m=100 cm

We have to find the wavelength of laser light.

We know that

\Delta y=\frac{\lambda L}{d}

Substitute the values

1.56\times 10^{-2}=\frac{\lambda\times 4.75}{0.23\times 10^{-3}}

\lambda=\frac{1.56\times 10^{-2}\times 0.23\times 10^{-3}}{4.75}

\lambda=7.55\times 10^{-7} m

4 0
3 years ago
What I Know Directions: Read and understand the questions below. Choose the letter of the best answer, Write the chosen letter o
frozen [14]
Meter is the basic unit of distance
3 0
2 years ago
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