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2 years ago

What happens to the light that strikes the Black Paint

Physics

2 answers:

SSSSS [86.1K]2 years ago

8 0

Answer:

It absorbs all the wavelengths of visual light that fall on it and no light is reflected into the eye from that object.

Hope this helps

Marysya12 [62]2 years ago

8 0

The object absorbs some amount of light and reflects back the rest.

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In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t

USPshnik [31]

Answer:

$W_x = 0.9156\ c$

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

$W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}$

u_x = 0.543 c

v_x = - 0.741 c

$W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}$

$W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}$

$W_x = \dfrac{1.284c}{1+0.402363}$

$W_x = 0.9156\ c$

Relative velocity of the particle is $W_x = 0.9156\ c$

5 0

3 years ago

The energy of an electromagnetic wave changes proportionally to which

ozzi

Answer:

D. Frequency

Explanation:

The energy of an electromagnetic wave is proportional to frequency, mathematically it is expressed as;

E ∝ f

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ Js

The equation above can also be expanded to;

$E = hf = h \frac{c}{\lambda}$

where;

c is speed of light = 3 x 10⁸ m/s

λ is the wavelength of the electromagnetic wave

Since the speed of light is constant, we can conclude that the energy of the electromagnetic wave is directly proportional to its frequency and inversely proportional to its wavelength.

Therefore, the correct option for direct proportionality is FREQUENCY

8 0

2 years ago

A goose waddles 18.0 m north and then 5.00 m east. He then walks 2.00 m south and 2.00 m west. What is the magnitude and directi

olya-2409 [2.1K]

Your answer should be 16.3 m 79.4º east of north

7 0

2 years ago

Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac

Serjik [45]

The strength of the gravitational field is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
$r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m$

And now we can use the previous equation to calculate the field strength at that altitude:
$g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2} = 8.95 m/s^2$

And we can see this value is a bit less than the gravitational strength at the surface, which is $g_s = 9.81 m/s^2$ .

4 0

3 years ago

Which is true of oxidation? PLS HELP FAST and thank you!

elena-14-01-66 [18.8K]

Answer:The Answer Is D

Explanation:

4 0

2 years ago