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Evgesh-ka [11]
3 years ago
9

I love you thanks for help❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️solveeeeee❤️❤️❤️❤️❤️❤️plzzzzz

Physics
2 answers:
Kipish [7]3 years ago
6 0

Answer:

okay

Explanation:

iogann1982 [59]3 years ago
5 0

Answer:

aaaaaaaaaaaaaaaa

Explanation:

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a student calculates experimentally the value of density of an iron as 4.4 gcm³. if the actual density of an iron is 7.6 gcm³, c
ser-zykov [4K]
Hola!

Percentage Error is a measurement of the discrepancy between an observed and a true, or accepted value.

[ refer the attachment. ]

According to Question,

% error = \frac{7.4 - 7.6}{7.6} × 100

= 2.631 % = 2.7 % (approximately.)

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3 years ago
Choose the false statement regarding resistance.
Afina-wow [57]

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The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be
Oksi-84 [34.3K]

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>
  • It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
  • Mathematically, T²∝a³
<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>
  • The time period of geosynchronous orbit is 24 hours or 1440 minutes.
  • As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
  • If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
  • a1= (T1/T2)⅔×a2

           = (1440/90)⅔×6780

           = 43,090 km

  • Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

brainly.com/question/16705471

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2 years ago
A pendulum on a grandfather clock
egoroff_w [7]
4.0 ilynits per second Alaskan es muy du facial in the oscillates 1.99
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3 years ago
The force F1, 10.0N acts at 10.0cm. What is the magnitude of the torque due to F1 about an axis through Point A perpendicular to
Nookie1986 [14]
Thank you for posting your question here at brainly. I hope the answer will help you. Below is the solution. Feel free to ask more question. 

<span>torque = rF
= 0.1(10)
=1 Nm</span>
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3 years ago
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