The Medical College - Page 64
glass prism toward air of index,
nair = 1.
Using
Snell's law:
nglass
sin Øc
= nair
sin 90°, and because nglass =
1.52; sin Øc = 1/1.52 and
Øc =
41°.
76.
D.
The objective lens is a projection lens,
pro-
ducing an inverted real image. The first
image
lies within the focal length of the eyepiece,
which
therefore acts as a magnifying glass, forming
a
virtual, enlarged image. The eyepiece does
not
invert the image.
77.
B.
As the object (primate) moves closer to
the
objective, the real image formed by the
objective
is further away from the objective, i.e.,
closer to
the original position of the eyepiece. To see
the
image clearly, the eyepiece must be
moved
"back" away from the objective in order to
focus
the second (virtual) image clearly. (It is the
eye-
piece that is adjustable in telescopes and
binocu-
lars.) This is similar to holding this printed
page
too close to your eyes to see clearly. If you
hold
the page still and move your head away you
will
be able to see the print clearly.
78.
B.
The pressure is due to the weight of
a
column of fluid 75 m tall. P = dgy; where d
=
density of the liquid, and y is the depth below
the
surface; g is the acceleration of gravity,
9.8
m/sec
2.
79.
C.
This is a simple conversion
problem:
1 atm
7.35
10 N/m
2
= 7.3 atm
1.01
10
5 n/m2
80.
C.
Pressure is force divided by area,
or
P = F/A; so F = PA. It is necessary to convert
the
area from cm2
to m2
(by multiplying by
10
-4 m2/cm2). Thus
F = 6.86
10
5 N/m2
65 cm
2
1
10
-4m2/cm2 = 4460 N.
81.
A.
Treating the air as an ideal gas, we see
that
P1V1 =
P2V2, or
V2 =
P1V1/P2. Notice
that we
must use the absolute pressures in this
expression.
Then P1
= 8.36
10
5 N/m2 and
P2 =
1.01
10
5
N/m
2.
82.
C.
The partial pressure is the same
fraction
(0.21) of the total absolute pressure of
8.36
10
5
N/m
2 as the fraction of
oxygen by volume in ordi-
nary air.
83.
C.
The work done on charged particles
by
electric fields depends on the charge of the
parti-
cle and the voltage difference (not on the mass
of
the particle). The electric work, W =
qV,
increases the kinetic energy of the particles,
in
this case from zero to the final KE. Because
the
charges for electron and proton are equal,
they
have equal final kinetic energies (although
the
electron has a greater speed).
84.
C.
We need to know the force (F = qvB
sin
)
on a moving charge in a magnetic field and
we
need the "right-hand rule" for the vector
direc-
tion of the force on a positively charged
particle
(knowing the vector velocity and vector
magnetic
field, B). For a positive charge moving
horizon-
tally north through a magnetic field pointing
hor-
izontally west, the magnetic force is
vertically
upward. The negative electron is equivalent to
a
positive charge moving in the opposite
direction
so the force on both electron and proton
is
upward.
85.
A.
The magnetic force acts only at right
angles
to the velocity of a charged particle. It
can
change the direction of the velocity vector
but
not its magnitude (speed is the magnitude
of
velocity).
86.
C.
The charged particles are now moving
par-
allel to the magnetic field direction so that
sin
is zero in the magnetic force equation (F =
qvB
sin
). The magnetic force on each is zero,
and
their paths are not deflected.
87.
B.
The electric force, F = qE, shows that
the
direction of the force is parallel to E for
positive
charges like protons, and opposite
(antiparallel)
for negative charges like
electrons.
88.
B.
The speed of any wave is the product of
its
frequency and wavelength, v = f
. This is true for
all waves, including sound, light, radio
waves,
and so forth. Thus
= v/f.
89.
A.
The angular acceleration is the change
in
angular velocity divided by the time taken,
or
=
(
f –
i)/t.
Converting
to rad/sec:
f = 57,300
rev/min
2
rad/rev
1 min/60sec = 6000 rad/sec
= (6000 rad/sec – 0)/120 sec = 50
rad/sec2
90.
B.
Under the conservation of energy
principle,
the initial kinetic energy will be entirely
converted
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