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3 years ago

Which example best illustrates that light behaves like particles?

Physics

2 answers:

AlekseyPX3 years ago

8 0

I would say B. Because actual mass would ricochet off the sidewalk.

mr_godi [17]3 years ago

8 0

Answer:

B. Light bounces off a white cement sidewalk.

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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

3 years ago

10. How much total work do you do when you lift a 50 kg microwave 1.0 m off the ground and then push it 1.0 m

Mariulka [41]

Work formula:

$W = Fd\cos(\theta)$

F = 50N, d = 1.0 m

When you lift something straight up, the angle of the force is 90º

cos(90º) is 0, so there's no work done when you lift the microwave off the ground

$W = (50N)(1.0)(0) = 0$

F = 50N, d = 1.0 m

When you push the microwave, the angle is 0º and cos(0º) is 1. So there is work done here:

$W = (50 N)(1.0m)(1)$

$W = 50$

total work = 50 joules

6 0

3 years ago

What is the relationship between the temperature of a star and its luminosity?

Sladkaya [172]

The Luminosity of a star is proportional to its Effective Temperature to the 4th power and its Radius squared.

3 0

3 years ago

A decrease in the magnitude of velocity is called

vagabundo [1.1K]

Any change in the speed or direction of motion is called "acceleration". You'll hear "deceleration" used for slowing down but that's not technically correct.

6 0

3 years ago

What does the prefix for kilo- mean

artcher [175]

The word/prefix " $kilo$ " means $one$ $thousad$

Good luck on your assignment and enjoy your day!

~ $MeIsKaitlyn :)$

3 0

3 years ago