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3 years ago

Isn't taking answers off this app cheating but it also helps with the good grade part.

Physics

2 answers:

Alex787 [66]3 years ago

7 0

Answer:

Period

Explanation:

It is not considered cheating its UseInG yOuR ReSoRcEs!

Yakvenalex [24]3 years ago

5 0

$Hello$ $There!$

Technically, Brainly is for kinda cheating... but, it does helps you get good grades in the future. :)

Hopefully, this helps you!!

$AnimeVines$

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Use Kepler’s laws and the period of the Moon (27.4 d) to determine the period of an artificial satellite orbiting very near the

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As according to Kepler 's law

T =(4π²r³/ Gm)^1/2

here r= distance from earth center to satellite = 6400km = 6400000m

G = earth's gravitational constant= 6.67×10^-11

m = mass of earth= 5.98 ×10^24

so T =[ { 4× (3.1416)²×(6400000)³}/ {(6.67×10^-11)×( 5.98 ×10^24)} ]^1/2

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3 years ago

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4 years ago

Highway transportation officials are trying to determine if the average speed on Archer road is above 45mph, the post speed limi

USPshnik [31]

Answer:

Option D) -0.0707

Explanation:

We are given the following in the question:

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Sample mean, $\bar{x}$ = 44.9

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Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

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3 years ago

What is the quantity of work done when a crane lifts a 100-n block from 2 m above the ground to 6 m above the ground?.

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Answer: by definition work = Force * distanceForce = 100Ndistance = 6 - 2 = 4work = 100 * 4 = 400 J or CStill stuck? Get 1-on-1 help from an expert tutor now.

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3 years ago

The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc

Mnenie [13.5K]

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

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Using formula of focal length

$f = \dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{1}{1.55}$

$f=0.645\ m$

$f=64.5\ cm$

We need to calculate the image distance

Using lens formula

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}$

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Hence, The image distance is 20.0 cm.

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3 years ago