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Rasek [7]
2 years ago
15

Relative formula mass of CuCO3

Physics
1 answer:
Elodia [21]2 years ago
5 0

AnMolar mass of CuCO3 = 123.5549 g/mol

This compound is also known as Copper(II) Carbonate.

Convert grams CuCO3 to moles  or  moles CuCO3 to grams

Molecular weight calculation:

63.546 + 12.0107 + 15.9994*3

Percent composition by element

Element   Symbol   Atomic Mass   # of Atoms   Mass Percent

Copper Cu 63.546 1 51.431%

Carbon C 12.0107 1 9.721%

Oxygen O 15.9994 3 38.848%

Explanation:

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A 600‑kg car accelerates at a rate of 3 m/s2. How much net force is acting on the car to cause this acceleration?
finlep [7]
600(kg) x 3(m/s^2) = 1800N (newtons)
4 0
3 years ago
A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax
Colt1911 [192]

Answer:

From the question we are told that

  The length of the rod is  L_o

    The  speed is  v  

     The angle made by the rod is  \theta

     

Generally the x-component of the rod's length is  

     L_x =  L_o cos (\theta )

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       L_xo  =  L_x  \sqrt{1  - \frac{v^2}{c^2} }

=>     L_xo  =  [L_o cos (\theta )]  \sqrt{1  - \frac{v^2}{c^2} }

Generally the y-component of the rods length  is mathematically represented as

      L_y  =  L_o  sin (\theta)

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e L_y

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}

=>  L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}

=>  L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}

=>   L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}

=> L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}

=> L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }

=> L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

Hence the length of the rod as measured by a stationary observer is

       L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

   Generally the angle made is mathematically represented

tan(\theta) =  \frac{L_y}{L_x}

=>  tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }

=> tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }

Explanation:

     

     

       

7 0
3 years ago
Three deer, a, b, and c, are grazing in a field. deer b is located 62.1 m from deer a at an angle of 54.3 ° north of west. deer
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by cosine law we know that

c^2 = a^2 + b^2 - 2 abcos\theta

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now using above equation

93.8^2 = 62.1^2 + b^2 - 2*62.1*b * cos46.6

4942.03 = b^2 - 85.34 b

b^2 - 85.34b - 4942.03 = 0

by solving above quadratic equation we have

b = 124.9 m

so it is at distance 124.9 m from deer a

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A kite suspended in the sky is flowing back and forth. Which type of friction is being described? static friction rolling fricti
expeople1 [14]
<span>fluid friction of the air also known as drag</span>
4 0
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Answer:

20.7 billion = 20.7E9        converting to units of 10

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69000 s = 1150 min = 19.2 hrs

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3 years ago
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