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Svetach [21]
3 years ago
15

To balance a chemical equation we use______. -coefficients -subscripts

Physics
1 answer:
Vitek1552 [10]3 years ago
3 0

Answer:

Coefficients

Explanation:

  • Coefficients are used to balance chemical equations.
  • According to the law of conservation of mass, the mass of the reactants in a chemical equation should be equivalent to the mass of the products.
  • The conservation of mass is achieved in chemical equations is achieved through balancing chemical equations.
  • <em><u>Balancing chemical equations is a try and error of putting appropriate coefficients to the reactants or products to ensure that the number of atoms of each element are equal on the side of the reactants and that of products. </u></em>
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When you jump,you push the earth downward.Why can’t you tell that the earth has moved?
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The movement of the earth is so small, especially compared to you, that you don't realize it. This is because the earth's mass is so much greater than your mass.

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3 years ago
A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The
lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

E_{i} = E_{f}

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}                              

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s      

 

3) When the object is at the equilibrium position, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s

I hope it helps you!                                                                                        

8 0
3 years ago
A 2 kg ball is thrown down with 50J of energy from a height of 10m, what is its velocity before it strikes the ground (neglect a
san4es73 [151]

Answer:

V=14

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V=14

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PLEASE HELP ME WITH THIS QUESTION!
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Answer: Top left to bottom right, top right to bottom left

Explanation:

A linear position time graph means a horizontal line on the velocity graph so top left matches the bottom right

With the process of elimination, bottom left matches the top right

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If there is no friction and no horizontal force acting on the bicycle, then the bicycle keeps rolling at a constant speed in a straight line, until the cows come home, Dante's Inferno freezes over, and the POTUS accepts some responsibility for his words, actions, and consequences.

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