Assignment 2b Solutions

  • June 2020
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Name: ______________________________Student no: _____________ Group:_______________ CC2404 Applied Physics and Instrumentation in Health Care Assignment 2b (2009) 1.

When a stimulus applied to a nerve cell, the cell is said depolarized. Which ion rushes into the inside cell. a. K+ b. Clc. Na+ d. K+ and ClAnswer____c_____

2.

(1 mark)

The resting potentials is about: a. -1 mV to -20mV b. -70mV to -90 mV c. +20 mV to +30 mV d. -20mV to -30 mV Answer____b_____

3.

(1 mark)

Which of the following has the correct meaning for a typical ECG waveform? a. P-wave : depolarization of ventricles b. QRS-wave: depolarization of ventricles c. P-wave : re-polarization of ventricles d. T-wave: depolarization of atria Answer____b____

(1 mark)

4. A parallel plate capacitor has a capacitance of 2 µ F is air and 4.6 µ Fwhen it is immersed in benzene. What is the dielectric constant of benzene? Solution  C = κεo

C1 κ1 A = ,∴ C2 κ 2 d

: in air κ is 1 , ∴κ 2 = κ1

(1 marks)

C2 4.6 = 1× = 2.3 C1 2

.

the dielectric constant of benzene is 2.3 (2 marks)

5. A 120 W light bulb is designed to operate on 220 V AC. (a)What is the rms current drawn by the bulb? (b)What is the peak current? (c) What is the resistance of the bulb’s filament? Pav 120 = = 0.545 A Solution: (a) I rms = Vrms 220 (b) I o = 2 I rms = 2 × 0.545 = 0.77 A

(1 mark) (1 mark)

Name: ______________________________Student no: _____________ Group:_______________ (c) R =

Vrms 220 V 2 rms 2202 = = 404 Ω or R = = = 404Ω I rms 0.545 Pav 120

(1 mark)

6. A negative hydrogen ion has an extra electron and a mass of 1.67 ×10 −27 kg . (a) Calculate its kinetic energy at 1.0% of speed of light (c= 3 ×10 8 m / s ) . (b) What acceleration voltage is needed to obtain this energy? ( q e = −1.6 ×10 −19 C ) Solution: a. K =

1 1 mv 2 = ×1.67 ×10 −27 × (0.01 × 3 ×10 8 ) 2 = 7.52 ×10 −15 J 2 2

(2 marks)

b. this kinetic energy is transformed by an electrical potential energy change ∆ EPE = QV, therefore : V =

∆EPE 7.52 ×10 −15 = = −46969 V = −46 .97 kV Q − 1.6 ×10 −19

(2 marks)

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