When you first pull back on the pendulum, and when you pull it back really high the Potential Energy is high and the Kinetic Energy is low, But when up let go, and it gets right around the middle, that's when the Potential energy transfers to Kinetic, at that point the kinetic Energy is high and the potential Energy is low. But when it comes back up at the end. The same thing will happen, the Potential Energy is high, and the Kinetic Energy is low. Through all of that the Mechanical Energy stays the same.
I hope this helps. :)
Answer:
<u>Yes</u>
Explanation:
Remember, <u>Newton's third law of motion;</u> which says in part that <em>"Every action has an equal and opposite reaction."</em>
Hence, in this case, the fact that the doorbell rang out implies that there was another force that was exerted on it; which is, John's finger pressing the doorbell.
In other words, when John uses his fingers to press the doorbell button he applies a force (a mechanical force), and that force results in an opposite reaction; the ringing of the doorbell.
<em>The statement that gives the relationship between energy needed in breaking a bond and the one that is released after breakin</em>g is
The amount of energy it takes to break a bond is always less than the amount of energy released when the bond is formed.
- Bond energy can be regarded as amount of energy that is required in breaking a particular bond.
- For a bond to be broken Energy will be added and when a bond is broken there will be release of energy
- Bond breaking can be regarded as endothermic process, it is regarded as endothermic because there is a lot of energy required to be absorbed.
- Where ever a bond is broken, there must be formation of another bond
- Bond forming on the other hand can be regarded as exothermic process, since there is a release of releases energy.
Therefore, more energy is required in breaking of bond compare to energy released after breaking of bond.
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Answer:
34g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
H2S + 2AgNO3 —> 2HNO3 + Ag2S
Next, we shall determine the number of mole of H2S required to react with 2 moles of AgNO3.
This is illustrated below:
From the balanced equation above,
We can see that 1 mole of H2S is required to react completely with 2 moles of AgNO3.
Finally, we shall convert 1 mole of H2S to grams. This is shown below:
Number of mole H2S = 1 mole
Molar mass of H2S = (2x1) + 32 = 34g/mol
Mass = number of mole x molar Mass
Mass of H2S = 1 x 34
Mass of H2S = 34g
Therefore, 34g of H2S is needed to react with 2 moles of AgNO3.
