Physics 3LC Exam Questions With Complete Solutions
(1 pt) A beam of microwaves with = 0.9 mm is incident upon a
12 cm slit. At a distance of 1 m from the slit, what is the
approximate width of the slit's image? Correct Answers 12cm
A bothersome feature of many physical measurements is the
presence of a background signal (commonly called "noise"). In
Part 2.2.4 of the experiment, some light that reflects off the
apparatus or from neighboring stations strikes the photometer
even when the direct beam is blocked. In addition, due to
electronic drifts, the photometer does not generally read 0.0 mV
even in a dark room. It is necessary, therefore, to subtract off
this background level from the data to obtain a valid
measurement. Suppose the measured background level is 4.5
mV. A signal of 19.7 mV is measured at a distance of 29.5 mm
and 16.6 mV is measured at 33.5 mm. Correct the data for
background and normalize the data to the maximum value. What
is the normalized corrected value at 33.5 mm? Correct Answers
19.7- 4.5 = 15.2 mV
16. 6- 4.5 = 12.1 mV
max = 15.2
12.1/15.2 = .79
A common technique in analysis of scientific data is
normalization. The purpose of normalizing data is to eliminate
irrelevant constants that can obscure the salient features of the
data. The goal of this experiment is to test the hypothesis that
the flux of light decreases as the square of the distance from the
source. In this case, the absolute value of the voltage measured
by the photometer is irrelevant; only the relative value conveys
,useful information. Suppose that in Part 2.2.2 of the experiment,
students obtain a signal value of 181 mV at a distance of 4.2 cm
and a value of 80 mV at a distance of 6.1 cm. Normalize the
students' data to the value obtained at 4.2 cm. (Divide the signal
value by 181.) Then calculate the theoretically expected
(normalized) value at 6.1 cm.
Normalized experimental value at 6.1 cm
Theoretically expected normalized value at 6.1 cm Correct
Answers 80/181mv = .44 = normalized experimental value
v1r1^2=v2r2^2
v2=v1(r1/r2)^2 = 1(4.2/6.1)^2 = .47 = theoretical
A reflected ultrasound pulse brings back two kinds of
information. The ___________ of the reflected pulse relative to
that of the incident pulse gives information on the type of
interface that produced the reflection. The _______________
gives information about the distance between the transducer and
the reflecting interface.
A. amplitude; travel time between when a pulse leaves and
when it returns to the transducer;
B. frequency; amplitude of the reflected pulse;
C. travel time; frequency of the reflected pulse;
D. None of the above Correct Answers Amplitude, travel time
between when a pulse leaves and when it returns to the
transducer.
,A typical sheet of weighing paper is 4 inches by 4.2 inches and
weighs 0.4 g. What is its absorber thickness in ? Correct
Answers convert inches to cm via 1in. = 2.54 cm
4 in * 2.54 = 10.16 cm
4.2 in * 2.54 = 9.91 cm
absorber thickness = (g)/(cm * cm) --> (0.4g)/(10.16*10.668) =
0.00369 g/cm^2
An alpha cannot penetrate your skin. Once inside the body,
however, an alpha source is the most hazardous radiation since it
causes the most ionizing events per unit length. There is really
only one way a sealed alpha source can hurt you. What is it?
A. If it comes from a uranium source
B. By ingestion
C. By direct contact with skin
D. None of the above Correct Answers B. By ingestion
An application of this principle is that a line mounted on
transparent slide casts the same diffraction pattern as a dark film
with a slot of equal size cut in it. In Part 6.2.5 of the experiment,
you will exploit this principle to measure the width of a hair. If
the distance between the first spot and the central minimum is s
= 1.6 cm, L = 11 m, and = 7 x m, what is the width of the hair?
Correct Answers w= λ*L/s
.48125mm
, An interface between air and some other substance, like tissue or
water, strongly scatters ultrasound waves. Which of the
following is NOT a consequence of this?
A. Images are sharper with higher ultrasound frequencies.
B. Sound transducers should be carefully coupled to the skin via
gel.
C. Lungs have high ultrasound attenuation rates.
D. All of the above are consequences. Correct Answers A.
Image are sharper with higher ultrasound frequencies.
An ordinary dental X-ray uses which imaging technique(s)?
A. Sonography
B. Projection
C. Tomography
D. None of the above Correct Answers Projection
As an example, mercury-203 is a radioisotope that is important
for kidney function studies and renal imaging. The physical half-
life, , is 46.6 days and biological half-life, , of is 10.3 days. What
is the effective half life of mercury-203?
days Correct Answers Teff = (TbTp)/(Tb+Tp)
Teff = (10.3 x 46.6)/(10.3 + 46.6)
A: 8.4355 days
(1 pt) A beam of microwaves with = 0.9 mm is incident upon a
12 cm slit. At a distance of 1 m from the slit, what is the
approximate width of the slit's image? Correct Answers 12cm
A bothersome feature of many physical measurements is the
presence of a background signal (commonly called "noise"). In
Part 2.2.4 of the experiment, some light that reflects off the
apparatus or from neighboring stations strikes the photometer
even when the direct beam is blocked. In addition, due to
electronic drifts, the photometer does not generally read 0.0 mV
even in a dark room. It is necessary, therefore, to subtract off
this background level from the data to obtain a valid
measurement. Suppose the measured background level is 4.5
mV. A signal of 19.7 mV is measured at a distance of 29.5 mm
and 16.6 mV is measured at 33.5 mm. Correct the data for
background and normalize the data to the maximum value. What
is the normalized corrected value at 33.5 mm? Correct Answers
19.7- 4.5 = 15.2 mV
16. 6- 4.5 = 12.1 mV
max = 15.2
12.1/15.2 = .79
A common technique in analysis of scientific data is
normalization. The purpose of normalizing data is to eliminate
irrelevant constants that can obscure the salient features of the
data. The goal of this experiment is to test the hypothesis that
the flux of light decreases as the square of the distance from the
source. In this case, the absolute value of the voltage measured
by the photometer is irrelevant; only the relative value conveys
,useful information. Suppose that in Part 2.2.2 of the experiment,
students obtain a signal value of 181 mV at a distance of 4.2 cm
and a value of 80 mV at a distance of 6.1 cm. Normalize the
students' data to the value obtained at 4.2 cm. (Divide the signal
value by 181.) Then calculate the theoretically expected
(normalized) value at 6.1 cm.
Normalized experimental value at 6.1 cm
Theoretically expected normalized value at 6.1 cm Correct
Answers 80/181mv = .44 = normalized experimental value
v1r1^2=v2r2^2
v2=v1(r1/r2)^2 = 1(4.2/6.1)^2 = .47 = theoretical
A reflected ultrasound pulse brings back two kinds of
information. The ___________ of the reflected pulse relative to
that of the incident pulse gives information on the type of
interface that produced the reflection. The _______________
gives information about the distance between the transducer and
the reflecting interface.
A. amplitude; travel time between when a pulse leaves and
when it returns to the transducer;
B. frequency; amplitude of the reflected pulse;
C. travel time; frequency of the reflected pulse;
D. None of the above Correct Answers Amplitude, travel time
between when a pulse leaves and when it returns to the
transducer.
,A typical sheet of weighing paper is 4 inches by 4.2 inches and
weighs 0.4 g. What is its absorber thickness in ? Correct
Answers convert inches to cm via 1in. = 2.54 cm
4 in * 2.54 = 10.16 cm
4.2 in * 2.54 = 9.91 cm
absorber thickness = (g)/(cm * cm) --> (0.4g)/(10.16*10.668) =
0.00369 g/cm^2
An alpha cannot penetrate your skin. Once inside the body,
however, an alpha source is the most hazardous radiation since it
causes the most ionizing events per unit length. There is really
only one way a sealed alpha source can hurt you. What is it?
A. If it comes from a uranium source
B. By ingestion
C. By direct contact with skin
D. None of the above Correct Answers B. By ingestion
An application of this principle is that a line mounted on
transparent slide casts the same diffraction pattern as a dark film
with a slot of equal size cut in it. In Part 6.2.5 of the experiment,
you will exploit this principle to measure the width of a hair. If
the distance between the first spot and the central minimum is s
= 1.6 cm, L = 11 m, and = 7 x m, what is the width of the hair?
Correct Answers w= λ*L/s
.48125mm
, An interface between air and some other substance, like tissue or
water, strongly scatters ultrasound waves. Which of the
following is NOT a consequence of this?
A. Images are sharper with higher ultrasound frequencies.
B. Sound transducers should be carefully coupled to the skin via
gel.
C. Lungs have high ultrasound attenuation rates.
D. All of the above are consequences. Correct Answers A.
Image are sharper with higher ultrasound frequencies.
An ordinary dental X-ray uses which imaging technique(s)?
A. Sonography
B. Projection
C. Tomography
D. None of the above Correct Answers Projection
As an example, mercury-203 is a radioisotope that is important
for kidney function studies and renal imaging. The physical half-
life, , is 46.6 days and biological half-life, , of is 10.3 days. What
is the effective half life of mercury-203?
days Correct Answers Teff = (TbTp)/(Tb+Tp)
Teff = (10.3 x 46.6)/(10.3 + 46.6)
A: 8.4355 days
