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Aleksandr-060686 [28]
2 years ago
15

Which scientists contributed to discovering the universal law of gravitation? Check all that apply. Tycho Brahe Albert Einstein

Johannes Kepler Nicolaus Copernicus Sir Isaac Newton Robert Hooke
Physics
2 answers:
stich3 [128]2 years ago
6 0
<span>Johannes Kepler
</span>Isaac Newton

<span>Scientists who contributed to discovering the universal law of gravitation are Isaac Newton and Johannes Kepler. While, historically speaking, the discovery of gravitation is given to Newton, he said that he based his work on the law of motion, discovered by Kepler.  </span>
jekas [21]2 years ago
4 0
Scientists who contributed to discovering the universal law of gravitation are isaac newton and Johannes Kepler. While, traditionally speaking, the invention of gravitation is given to Newton, the same that he based mostly his work on the law of motion, discovered by <span>Johan Kepler.

Explanation:
</span>Sir Isaac<span> Newton: The Universal Law of Gravitation. </span>there's a well-liked<span> story that Newton was sitting </span>underneath a fruit tree<span>, </span>an apple<span> fell on his head, and he suddenly thought of the Universal Law of Gravitation</span>

Kepler's laws explain how the planets moved around the sun but not why. Newton filled in that gap by supposing there was a force acting between the bodies that were moving around each other.

The story goes that Newton saw an apple fall to the ground and it made him wonder why the fruit always fell straight to the ground; why did it not veer off to the left or right? According to his own laws of motion, anything that begins moving from a standing start is undergoing acceleration and, where there is acceleration, there must be a force.

 The apple started in the tree and landed on the Earth, which means there must be a force of attraction between the apple and the Earth.


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Which statements describe the applications of nuclear medicine scans? Check all that apply.
coldgirl [10]

Answer:

- asses disease progression and tissue function

- utilize a biologically active molecule

- utilize a radionuclide tracer

Explanation:

4 0
2 years ago
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The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th
Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em>1.83\times 10^{7}\,kg<em>. The wheels of the locomotive push back on the tracks with a constant net force of </em>7.50\times 10^{5}\,N<em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force (F), measured in newtons, then acceleration (a), measured in meters per square second, must be constant and can be found by the following expression:

a = \frac{F}{m} (1)

Where m is the mass of the freight train, measured in kilograms.

If we know that F = 7.50\times 10^{5}\,N and m = 1.83\times 10^{7}\,kg, then the acceleration experimented by the train is:

a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}

a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}

Now, the time taken to accelerate the freight train from rest (t), measured in seconds, is determined by the following formula:

t = \frac{v-v_{o}}{a} (2)

Where:

v - Final speed of the train, measured in meters per second.

v_{o} - Initial speed of the train, measured in meters per second.

If we know that a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}, v_{o} = 0\,\frac{m}{s} and v = 22.222\,\frac{m}{s}, the time taken by the freight train is:

t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s}  }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }

t = 542.265\,s

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0
2 years ago
A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitu
Mumz [18]

Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

T=\dfrac{2\pi}{\omega}

0.74=\dfrac{2\pi}{\omega}

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude  = 0.13 m

8 0
2 years ago
Astronomers initially had difficulty identifying the emission lines in quasar spectra at optical wavelengths because
Rus_ich [418]
No one expected violet & ultraviolet spectral lines to be shifted towards the red.
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2 years ago
A physics student with too much free time drops a watermelon from a roof of a building, hears the sound of the watermelon going
tatiyna

Answer:

28.6260196842 m

Explanation:

Let h be the height of the building

t = Time taken by the watermelon to fall to the ground

Time taken to hear the sound is 2.5 seconds

Time taken by the sound to travel the height of the cliff = 2.5-t

Speed of sound in air = 340 m/s

For the watermelon falling

s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow h=\frac{1}{2}\times 9.81\times t^2

For the sound

Distance = Speed × Time

\text{Distance}=340\times (2.5-t)

Here, distance traveled by the stone and sound is equal

\frac{1}{2}\times 9.81\times t^2=340\times (2.5-t)\\\Rightarrow 4.905t^2=340\times (2.5-t)\\\Rightarrow t^2=\frac{340}{4.905}(2.5-t)\\\Rightarrow t^2+69.3170234455t-173.292558614=0

t=\frac{-69.31702\dots +\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1},\:t=\frac{-69.31702\dots -\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1}\\\Rightarrow t=2.4158\ s\ or\ -71\ seconds

The time taken to fall down is 2.4158 seconds

h=\frac{1}{2}\times 9.81\times 2.4158^2=28.6260196842\ m

Height of the buidling is 28.6260196842 m

7 0
2 years ago
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