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Gala2k [10]
3 years ago
5

How fast is the earth moving through the universe?

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
8 0
The Earth moves 30 km/s in it's orbit around the sun.
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When you attract every object in the universe with a force that is proportional to the mass of the objects and to the distance b
yuradex [85]

When you attract every object in the universe with a force that is proportional to the mass of the objects and to the distance between them, we are obeying Newton's law of universal gravitation.

<h3>Newton's law of universal gravitation</h3>

Newton's law of universal gravitation states that the force of attraction between two masses in the universe is directly proportional to the product of the masses and inversely proportional to the the square of the distance between them.

The mathematical interpretation of the above law is

  • F ∝Mm/r²

Removing the proportionality sign,

  • F = GMm/r².

Where:

  • F = Force of attraction
  • G = Gravitational constant
  • M = Bigger mass
  • m = Smaller mass
  • r = Distance between the masses.

From the above, When you attract every object in the universe with a force that is proportional to the mass of the objects and to the distance between them, we are obeying Newton's law of universal gravitation.

Learn more about Newton's law of universal gravitation here: brainly.com/question/9373839

#SPJ12

7 0
2 years ago
John had a stroke and is having difficulty sitting and walking. Which medical professional should he consult to improve his phys
slava [35]
John needs to see a physical therapist because he cannot walk very well.
5 0
3 years ago
Read 2 more answers
A light ray moving at a 54.3 deg
MrMuchimi

Answer:

45.25^{\circ}

Explanation:

Applying Snell's law

n_1sin\theta_1 = n2*sin\theta_2\\1.33*sin(54.3) = 1.52*sin\theta_2\\sin\theta_2 = 1.33*0.81208353/1.52 = 0.71057\\\theta_2=sin^{-1} (0.71057)\\\theta_2 = 45.25^{\circ}

3 0
3 years ago
Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3
VARVARA [1.3K]

Answer:

Part a)

f = 4.76 \times 10^{14} Hz

Part b)

d = 3.48 \times 10^{-4} m

Part c)

\theta = 0.311 degree

Explanation:

Part a)

As we know that the speed of light is given as

c = 3 \times 10^8 m/s

\lambda = 630 nm

now the frequency of the light is given as

f = \frac{c}{\lambda}

so we have

f = \frac{3 \times 10^8}{630 \times 10^{-9}}

f = 4.76 \times 10^{14} Hz

Part b)

Position of Nth maximum intensity on the screen is given as

y_n = \frac{n\lambda L}{d}

so here we know for 3rd order maximum intensity

y_3 = 0.76 cm

n = 3

L = 1.4 m

0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}

d = 3.48 \times 10^{-4} m

Part c)

angle of third order maximum is given as

d sin\theta = 3 \lambda

3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})

\theta = 0.311 degree

8 0
3 years ago
How fast would a 2000kg car need to be traveling to have the same momentum has a 3000kg truck driving at 18m/s?
Marrrta [24]
p = mv = 3000  * 18 = 2000 * x

solve for x.
5 0
3 years ago
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