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pickupchik [31]
3 years ago
10

Kepler's laws follow which law discovered by Sir Isaac Newton?

Physics
2 answers:
11111nata11111 [884]3 years ago
4 0

B. Universal Gravitational Law

katrin [286]3 years ago
4 0

Answer:

B. Universal Gravitational Law

Explanation:

As we know that Kepler gives three laws regarding the planetary motion of all planets around the sun

All his laws are derived from the basic Theory of law of attraction by gravitation.

Universal law of attraction given by Sir Issac Newton says that if two masses are placed at some distance from each other then the net force between two masses is given as

F = \frac{Gm_1 m_2}{r^2}

here we know that

m_1, m_2 = masses of two objects

r = distance between two objects

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It depends on your weight, your hieght, and how fast you are falling
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Do you think radio waves and x-rays are types of light?
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A starship travels to a planet that is 20 light years away. The astronauts stay on the planet for 2.0 years before returning at
ad-work [718]

Answer:

astronauts age is 32 years

correct option is e 32 years

Explanation:

given data

travels = 20 light year

stay = 2 year

return = 52 years

to find out

astronauts aged

solution

we know here they stay 2 year so time taken in traveling is

time in traveling = ( 52 -2 )  = 50 year

so it mean 25 year in going and 25 years in return

and distance is given 20 light year

so speed will be

speed = distance / time

speed = 20 / 25 = 0.8 light year

so time is

time = \frac{t}{\sqrt{1-v^2} }

time =  \frac{25}{\sqrt{1-0.8^2} }

time = 15 year

so age is 15 + 2 + 15

so astronauts age is 32 years

so correct option is e 32 years

4 0
3 years ago
The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 440g . The vibra
ANEK [815]

Answer:

Tension, T = 2038.09 N

Explanation:

Given that,

Frequency of the lowest note on a grand piano, f = 27.5 Hz

Length of the string, l = 2 m

Mass of the string, m = 440 g = 0.44 kg

Length of the vibrating section of the string is, L = 1.75 m

The frequency of the vibrating string in terms of tension is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

\mu=\dfrac{m}{l}

\mu=\dfrac{0.44}{2}=0.22\ kg/m

T=4L^2f\mu

T=4\times (1.75)^2\times (27.5)^2 \times 0.22

T = 2038.09 N

So, the tension in the string is 2038.09 N. Hence, this is the required solution.

6 0
3 years ago
The drawing shows a skateboarder moving at 4.8 m/s along a horizontal section of a track that is slanted upward by 48° above the
Anna11 [10]
Let M = mass of the skier, 
v2 = his speed at the end of the track. 
By conservation of energy, 
1/2 Mv^2 = 1/2 Mv2^2 + Mgh 
Dividing by M, 
1/2 v^2 = 1/2 v2^2 + gh
 Multiplying by 2, 
v^2 = v2^2 + 2gh 
Or v2^2 = v^2 - 2gh 
Or v2^2 = 4.8^2 - 2 * 9.8 * 0.46 
Or v2^2 = 23.04 - 9.016 
Or v2^2 = 14.024 m^2/s^2-----------------------------(1) 
In projectile motion, launch speed = v2 
and launch angle theta = 48 deg 
Maximum height 
H = v2^2 sin^2(theta)/(2g) 
Substituting theta = 48 deg and value of v2^2 from (1),
 H = 14.024 * sin^2(48 deg)/(2 * 9.8) 
Or H = 14.024 * 0.7431^2/19.6 
Or H = 14.024 * 0.5523/19.6 
Or H = 0.395 m = 0.4 m after rounding off 
Ans: 0.4 m

The answer in this question is 0.4 m
4 0
3 years ago
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