Answer:
Yes, this is completely independent.
Explanation:
Yes, this is completely independent. Even though there are no South American individuals that are majoring in biomedical engineering in this party it is still a completely independent factor. The origin of birth of an individual does not tie them to a specific degree or field of expertise, therefore a South American individual can study anything they want including mechanical engineering, electrical engineering, or biomedical engineering.
(a) At a simple interest rate of 12% per year, determine how long it will take $5000 to increase to twice as much. (b) Compare the time it will take to double if the rate is 20% per year simple interest.
Explanation:
10000=5000(1.12^x)
2=1.12^x
(log_1.12)(2)=x
x= about 6.1163
10000=5000(1.2^x)
2=1.2^x
(log_1.2)(2)=x
x= about 3.8019
compare them by saying like 20% will be 6.12/3.8 times faster
A closed system of mass 20 kg undergoes a process in which there is a heat transfer of 1000 Q6: ki from the system to the surroundings. The work done on the system is 200 kl. If the initial 5 specific internal energy of the system is (250+R:) kl/kg, what is the final specific internal energy, in kj/kg? Neglect changes in kinetic and potential energy:
The final specific internal energy : 190 kJ/kg
Further explanationThe laws of thermodynamics 1 state that: energy can be changed but cannot be destroyed or created
The equation is:
[tex]\tt E_{in}-E_{out}=\Delta E~system\\\\\Delta E=\Delta U+\Delta KE+\Delta PE\\\\\Delta U=m(U_2-U_1)\\\\Q-W=\Delta U+\Delta KE+\Delta PE[/tex]
Energy owned by the system is expressed as internal energy (U)
This internal energy can change if it absorbs heat Q (U> 0), or releases heat (U <0). Or the internal energy can change if the system does work or accepts work (W)
The sign rules for heat and work are set as follows:
• The system receives heat, Q +
• The system releases heat, Q -
• The system does work, W -
• the system accepts work, W +
A closed system of mass 20 kg⇒m=20 kg
Heat transfer of 1000 kJ from the system to the surroundings⇒Q=-1000 kJ
The work done on the system is 200 kJ⇒W=+200 kJ
The initial specific internal energy of the system is 250 kJ /kg⇒U₁ = 250 kj/kg
Neglect changes in kinetic and potential energy⇒ΔKE+ΔPE=0, so
Q-W = ΔU
Input in equation
[tex]\tt -1000-200=20(U_2-250)\\\\-1200=20U_2-5000\\\\3800=20U_2\\\\U_2=190~kJ/kg[/tex]