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Papessa [141]
3 years ago
14

Which of the following elements is commonly found in the Earth's crust, living matter, oceans, and atmosphere? A. carbon B. argo

n C. bromine D. zinc
Physics
2 answers:
jekas [21]3 years ago
8 0
The answer is carbon.
OverLord2011 [107]3 years ago
5 0
I'm pretty sure it's A. carbon
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A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 34 m/s when a 60
german

Answer:

34 m/s

Explanation:

m = Mass of glider with person = 680 kg

v = Velocity of glider with person = 34 m/s

m_1 = Mass of glider without person = 680-60 kg

v_1 = Gliders speed just after the skydiver lets go

m_2 = Mass of person = 60 kg

v_2 = Velcotiy of person = 34 m/s

As the linear momentum of the system is conserved

m_1v_1+m_2v_2=mv\\\Rightarrow v_1=\dfrac{mv-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{680\times 34-60\times 34}{680-60}\\\Rightarrow v_1=34\ m/s

The gliders speed just after the skydiver lets go is 34 m/s

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3 years ago
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Andreyy89
It is important to maintain septic systems in order to <span>prevent untreated sewage from contaminating groundwater</span>
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3 years ago
A 20-N force acts on an object with a mass of 2.0kg. what is the objects acceleration?
ale4655 [162]

Answer:

10 m/s^2

Explanation:

Equation: F = ma.

a = acceleration

m = mass

F = force

Because we are trying to find acceleration instead of force we want to rearrange the equation to solve for a which is F/m = a.

F = 20

m = 2

a = ?

a = F/m

a = 20/2

a = 10 m/s^2

7 0
3 years ago
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Suppose you analyze standardized test results for a country and discover almost identical distributions of physics scores for fe
zaharov [31]

Answer:

go ask a teacher

Explanation:

go to school

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finish question done :)

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3 years ago
At a distance of 0.208 cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485 n/c . wh
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We have the equation for electric field E = kQ/d^{2}

Where k is a constant, Q is the charge of source and d is the distance from center.

In this case E is inversely proportional to d^{2}

So, \frac{E_{1} }{E_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}

E_{1} = 485 N/C

d_{1} = 0.208 cm

d_{2} = 0.620 cm

E_{2} = ?

\frac{485 }{E_{2}} = \frac{0.620^{2}}{0.208^{2}}

E_{2} = \frac{485*0.208^{2}}{0.628^{2}}

E_{2} = 53.20 N/C

7 0
3 years ago
Read 2 more answers
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