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maksim [4K]
3 years ago
11

Which metal in Period 5 is very reactive and has two valence electrons in each atom?

Physics
1 answer:
iren2701 [21]3 years ago
8 0
The answer is Strontium(Sr)
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Tell the value of the underlined digit 843,208,732,833 eight is underlined
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The value of the underlined 8 is, hundred billion's. Hope this helped!
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3 years ago
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A crane used 250,000 Joules of work to move a beam to the top of a building in 20 seconds. How much power did the crane use?
maria [59]

Answer:

12500W

Explanation:

Given parameters:

Work done  = 250000J

Time taken  = 20s

Unknown:

Power of the crane = ?

Solution:

Power is the defined as the rate at which work is being done;

  Mathematically;

        Power = \frac{work done}{time }

 insert the parameters and solve;

      Power  = \frac{250000}{20}   = 12500W

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3 years ago
Evaluate x and y in the equation: E=Cm^xV^y , where E is kinetic energy , m is mass , V is velocity and C is a dimension less co
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Answer: Do I look like Einstein

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2 years ago
A ship maneuvers to within 2500 m of an island's 1800 m high mountain peak and fires a projectile at an enemy ship 610 m on the
Ne4ueva [31]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

t=(0-(250sin75)^2)/-9.8 
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<span>the max height one is d=0.5*9.8*t^2 </span>
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3 0
3 years ago
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Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o
Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, M=1.99\times 10^{30}\ kg

Radius of Mercury's orbit, r=5.79\times 10^{10}\ m

Radius of discovered planet, R=\dfrac{2}{3}r

R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

T^2\propto R^3

T^2=\dfrac{4\pi^2R^3}{GM}

T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}

T=\sqrt{1.71\times 10^{13}}

T = 4135214.625 s

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T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

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