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Sladkaya [172]
3 years ago
6

What is the behavior that takes place when two waves of equal amplitude collide abd sum up to a bigger wave.

Physics
1 answer:
olga nikolaevna [1]3 years ago
3 0

That's "constructive interference".

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Place a small object on the number line below at the position marked zero. Draw a circle around the object. Mark the center of t
Viefleur [7K]
It’s is D bc it is and idk you should do it
6 0
3 years ago
I have about 4 questions can anyone help? Its about motion (I don't quite know what type of science that is...)
Neko [114]
Motion is physics. motion is the displacement,distance,velocity, acceleration, time, and speed.
6 0
3 years ago
On an asteroid, the density of dust particles at a height of 3 cm is ~30% of its value just above the surface of the asteroid. A
Anvisha [2.4K]

From the law of atmosphere

N_v(y) = n_0*e^{-\frac{mgy}{Kb*T}}

Where

n_0 = constant and is number density where the height y = 0cm

n_V = Number density at height y=3cm

Kb = Boltzmann constant = 1.38*10^{-23}J/K

T=20K

m = 10^{-19}kg

Re-arranging the equation to have the value of the gravity,

\frac{N_v(y)}{n_0} = e^{-\frac{mghy}{KbT}}

ln(\frac{N_v(y)}{n_0}) = -\frac{mgy}{KbT}

Since it is 30% of value above surface, therefore N_v = 0.3n_0

ln(\frac{0.3n_0}{n_0}) = -\frac{mgy}{KbT}

g = -\frac{KbT ln(0.3)}{my}

g = -\frac{(1.38*10^{-23}J/K)(20K)(Ln(0.3))}{10^{-19}(3*10^{-2})}

g = \frac{1.38*2*ln(0.3)*10^{-22}}{3*10^{-4}}

g = 1.104*10^{-1}m/s^2

g = 0.1m/s^2

Therefore the correct answer is C.

4 0
4 years ago
A light ray bounces off a fish in your aquarium. It travels
frutty [35]

Answer:

Correct

Explanation:

Because the light can't travel as quickly in the water as it does in the air, the light bends around the pencil, causing it to look bent in the water. Basically, the light refraction gives the pencil a slight magnifying effect, which makes the angle appear bigger than it actually is, causing the pencil to look crooked.

7 0
3 years ago
Read 2 more answers
an astronaut weighs 8.00x10^2 on the surface of earth. what is the weight of the astronaut 6.37x10^6 meters above the surface of
Marta_Voda [28]
To solve this problem, we use the Law of Universal Gravitation:

F = Gm1m2/d^2

where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:

F = k/d^2
k = Fd^2
F_{1} d_{1} ^{2} =F_{2} d_{2} ^{2}

The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1

(8E+2)(d1)^2 = F2(2d1)^2

(8E+2)(d1)^2 = 4F2(d1)^2

(8E+2)=4F2
F2 = 200 Newtons
4 0
3 years ago
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