Answer:
36.55 J
Explanation:
PE = Potential energy
KE = Kinetic energy
TE = Total energy
The following data were obtained from the question:
Position >> PE >>>>> KE >>>>>> TE
1 >>>>>>>> 72.26 >> 27.74 >>>> 100
2 >>>>>>>> 63.45 >> x >>>>>>>> 100
3 >>>>>>>> 58.09 >> 41.91 >>>>> 100
The kinetic energy of the pendulum at position 2 can be obtained as follow:
From the table above, at position 2,
Potential energy (PE) = 63.45 J
Kinetic energy (KE) = unknown = x
Total energy (TE) = 100 J
TE = PE + KE
100 = 63.45 + x
Collect like terms
100 – 63.45 = x
x = 36.55 J
Thus, the kinetic energy of the pendulum at position 2 is 36.55 J.
Answer:
43.05 moles of Al needed to react with 28.7 moles of FeO.
Explanation:
Given data:
Moles of FeO = 28.7 mol
Moles of Al needed to react with FeO = ?
Solution:
Chemical equation:
2Al + 3FeO → 3Fe + Al₂O₃
Now we will compare the moles of Al with FeO.
FeO : Al
2 : 3
28.7 : 3/2×28.7 = 43.05 mol
Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.
A cold front is the leading edge of a cooler mass of air, replacing at ground level a warmer mass of air, which lies within a fairly sharp surface trough of low pressure.
Voltage is known to be used as a requirement for stimulating an axon. The axon conducts electrical signals away from the nerve cell. The main function of the axons is that it is specifically used to transmit information, in a form of electrical impulse, to the other parts of the body.
Answer:
14.434 r.a.m.
Explanation:
- The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.
∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)
atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.
atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.
