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Neporo4naja [7]
2 years ago
6

Kelli and Jarvis are members of the stage crew for the Variety Show. Between acts, they must quickly move a Baby Grand Piano ont

o stage. After the curtain closes, they exert a sudden forward force of 524 N to budge the piano from rest and get it up to speed. The 158-kg piano experiences 418 N of friction. a. What is the piano's acceleration during this phase of its motion? b. If Kelli and Jarvis maintain this forward force for 1.44 seconds, then what speed will the piano have?
Chemistry
1 answer:
Marianna [84]2 years ago
3 0

If Kelli and Jarvis maintain a forward force of 524 N for 1.44 seconds on a 158-kg piano, the acceleration will be 0.671 m/s² and the final velocity will be 0.966 m/s.

Kelli and Jarvis exert a force of 524 N on a 158- kg (m) piano.

The friction force, which opposes this force, is 418 N.

The net force (F) is the difference between both forces.

F = 524 N - 418 N = 106 N

We can calculate the acceleration (a) of the piano using Newton's second law of motion.

<h3>What is Newton's second law of motion?</h3>

Newton's second law of motion states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object.

a = F/m = 106 N/158kg = 0.671 m/s²

The piano start from rest (u = 0 m/s) and moves with an acceleration of 0.671 m/s² for 1.44 s.

We can calculate its final velocity (v) using the following kinematic equation.

v = u + a × t

v = 0 m/s + 0.671 m/s² × 1.44 s = 0.966 m/s

If Kelli and Jarvis maintain a forward force of 524 N for 1.44 seconds on a 158-kg piano, the acceleration will be 0.671 m/s² and the final velocity will be 0.966 m/s.

Learn more about Newton's second law of motion here: brainly.com/question/10673278

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Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

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\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

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