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Olin [163]
3 years ago
6

Identify the energy transformations in the image below:

Physics
1 answer:
zvonat [6]3 years ago
4 0

Answer:

B

Explanation:

Electrical to Radiant.

I think

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In Part 6.2.2, you will determine the wavelength of the laser by shining the laser beam on a "diffraction grating", a set of reg
harkovskaia [24]

Answer:

λ = 2042 nm

Explanation:

given data

screen distance d = 11 m

spot s = 4.5 cm = 4.5 ×10^{-2} m

separation L = 0.5 mm = 0.5 ×10^{-3} m

to find out

what is λ

solution

we will find first angle between first max and central bright

that is tan θ = s/d

tan θ = 4.5 ×10^{-2}  / 11

θ = 0.234

and we know diffraction grating for max

L sinθ  = mλ

here we know m = 1  so put all value and find λ

L sinθ  = mλ

0.5 ×10^{-3}  sin(0.234)  = 1 λ

λ = 2042.02 ×10^{-9}  m

λ = 2042 nm

3 0
3 years ago
A ball is dropped off the roof of a tall building. If the ball reaches the ground in 8 seconds, how tall is the building, in met
Ghella [55]

Let <em>b</em> be the height of the building, and <em>y</em> the height of the ball at time <em>t</em>, given by

<em>y</em> = <em>b</em> - 1/2 <em>gt</em>²

where <em>g</em> = 9.8 m/s² is the magnitude of the acceleration due to gravity.

It takes the ball 8 s to reach the ground, at which point <em>y</em> = 0, so that

0 = <em>b</em> - 1/2 (9.8 m/s²) (8 s)²

<em>b</em> = 1/2 (9.8 m/s²) (8 s)²

<em>b</em> = 313.6 m

5 0
3 years ago
A horse runs a distance of 240 m in 20 s. Which of the following is a scalar quantity that can be determined from this
mariarad [96]

Explanation:

distance and time both are scaler quantity

6 0
3 years ago
Two racecars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the
Alenkasestr [34]

Answer:

The acceleration of car 2 is four times of the acceleration of car 1.

Explanation:

The centripetal acceleration of the object is possessed when it moves in a circular path. It is given by :

a=\dfrac{v^2}{r}

In this case, two race cars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the track, but car 2 is driving twice as fast as car 1.

So,

\dfrac{a_1}{a_2}=\dfrac{v_1^2}{v_2^2}

1 and 2 represent car 1 and car 2 respectively.

v_2=2v_1

So,

\dfrac{a_1}{a_2}=\dfrac{v_1^2}{(2v_1)^2}\\\\\dfrac{a_1}{a_2}=\dfrac{v_1^2}{4v_1^2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{4}\\\\a_2=4\times a_1

So, the acceleration of car 2 is four times of the acceleration of car 1.

4 0
3 years ago
Calculate the wavelength of 26 GeV electrons. Express your answer using two significant figures. λ = m
Alenkinab [10]

Answer:

4.78 x 10^-17 m

Explanation:

E = 26 GeV

First convert GeV into J

1 GeV = 1.6 x 10^-10 J

E = 26 x 1.6 x 10^-10 = 41.6 x 10^-10 J

Use the formula for energy

E = h c / λ

Where, h is the Plank's constant and λ be the wavelength and c be the velocity of light.

λ = h c / E

λ = ( 6.63 x 10^-34 x 3 x 10^8) / (41.6 x 10^-10)

λ = 4.78 x 10^-17 m  

6 0
3 years ago
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