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3 years ago

PLEASE ANSWER, I HAVE 5 MINUTES!!!!

Physics

1 answer:

zmey [24]3 years ago

3 0

Answer:

<em>I HAD THE SAME QUESTION AND SEARCHED IT UP AND GOT THE ANSWERS.</em>

Explanation:

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What chemical formula did early instigators assume for water?

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Answer:

Explanation:

Investigators thought this meant that oxygen was eight times more massive than hydrogen. They presumed the chem-ical form-ula for water to be H-O

Hope this helped!!! .

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3 years ago

When the force on an object and the object’s displacement are in different directions, which component of the force does work?

podryga [215]

Answer:

It should be b hope it helps :)

Explanation:

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3 years ago

Scientists use models to

xxTIMURxx [149]

D Is Thw Correct answer

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4 years ago

In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of

antoniya [11.8K]

Answer : 2446 years.

Explanation :

Length of semi major axis is, $a=84\ au= 1.496\times 10^{11}\ m$

According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.

i.e $T^2=\dfrac{4\pi^2}{GM}a^3$

where G is gravitational constant

M is mass of sun, $M=1.98\times 10^{30}\ Kg$

So, $T^2=\dfrac{4\times (3.14)^2}{6.6\times 10^{-11}Nm^2/Kg\times 1.98\times 10^{30}\Kg}$

$T^2=3\times 10^{-19}\times(84\times 1.496\times 10^{11})^3$

$T^2=3\times 10^{-19}\times 1984415.6\times 10^{33}$

$T^2=59532469.8\times 10^{14}\ s$

$T=7715.7\times 10^7\ s$

since, $1\ sec=3.17\times 10^{-8}\ years$

So, orbital period is approximately 2446 years.

7 0

4 years ago

Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,

seropon [69]

Answer:

The angular frequencies of all the 3 pendulums shall be same.

Explanation:

The time period of a simple pendulum with the approximation $sin(\theta )\approx \theta$ is given by:

$T=2\pi\sqrt{\frac{l}{g}}$

The angular frequency $\omega$ is given by

$\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}$

As we can see that the angular frequency is independent on the initial angle (valid strictly for small angle approximations) we conclude that the angular frequencies of the 3 pendulums are the same.

6 0

4 years ago