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yawa3891 [41]
3 years ago
15

The diagram shows parts of a wave. A series of waves with an arrow passing through their centers. The highest point of one wave

is labeled X. The lowest point of another wave is labeled Y. A bracket labeled Z connects the highest points of 2 waves. Which is shown by the letter X? trough amplitude crest wavelength
Physics
2 answers:
Alex Ar [27]3 years ago
7 0

Answer:

The answer is A "trough"

Explanation:

A

Pachacha [2.7K]3 years ago
3 0

Answer:

im not sure but i think its c

Explanation:

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The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i
spayn [35]

Answer:

<em>a) 3.56 x 10^22 N</em>

<em>b) 3.56 x 10^22 N</em>

<em></em>

Explanation:

Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = \frac{GMm}{R^2}

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = \frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2} = <em>3.56 x 10^22 N</em>

<em></em>

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to <em>3.56 x 10^22 N</em>

b) The force exerted by the Earth on the Sun = <em>3.56 x 10^22 N</em>

7 0
3 years ago
3
Elanso [62]

Answer:

ive got no idea just ask your teacher

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What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1
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Answer:

Explanation:

just use the gravational force equation which is G x m of earth x m of object divided by r squared (which is radius of earth)

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<span>Mass of the block m = 3.3kg Angle of the slide = 30 degrees Distance the block slides s = 2.10 m Time taken to slide t = 1.6 s Initially in rest condition so initial velocity u = 0. We have an equation for distance s = (u x t) + (1/2) x (a t^2) s = (0 x t) + (1/2) x (a x (1.6) ^2) => 2.10 = (1/2) x (a x2.56) 2.56a = 4.20 => a = 1.64 So the magnitude of the Acceleration a = 1.64 m/s^2</span>
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