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Sonbull [250]
3 years ago
14

Which number is the same as 3.11 x 10-4 ?

Physics
2 answers:
andrew-mc [135]3 years ago
7 0

Answer:

number is same as 0.000311

Explanation:

As we know that the number in scientific representation is given as

N = 3.11 \times 10^{-4}

also we know that

now we know that

10^{-4} = \frac{1}{10^4}

10^{-4} = \frac{1}{10000}

now we have

3.11 \times 10^{-4} = \frac{3.11}{10000}

3.11 \times 10^{-4} = 0.000311

Arturiano [62]3 years ago
6 0
10^-4 = 0.0001 * 3.11 = 0.000311
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If the period of oscillation of a simple pendulum is 4s, find its length. If the velocity of the bob
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Answer:

Explanation:

Because we assume the pendulum is a "mathematical pendulum" (neglecting the moment of inertia of the bob), we can find:

T=2\pi\sqrt{\frac{L}{g}} \rightarrow 4=2\pi\sqrt{\frac{L}{9.81}} \rightarrow \frac{4}{\pi^{2}}=\frac{L}{9.81} \rightarrow L \approx 3.97 m

By using the y=A\sin(\omega t)  \rightarrow v = \frac{dy}{dt}=\omega A \cos\omega t = \omega\sqrt{A^{2}-y^{2}}

The mean position is the position when <em>y</em> = 0, so:

\omega = \frac{2\pi}{T}=\frac{2\pi}{4}=0.5\pi rad/s

and v = \omega A \rightarrow A=\frac{40}{0.5\pi}=\frac{80}{\pi} in centimeters (cm).

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2 years ago
Need help with some physics questions
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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

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  • 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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A force of 10 N causes a spring to extend by 20 mm. Find: a) the spring constant of the spring in N/m​
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Answer:

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