DATA The statistical quantities “average value” and “root-mean-square value” can be applied to any distribution. Figure P18.82 shows the scores of a class of 150 students on a 100-point quiz. (a) Find the average score for the class. (b) Find the rms score for the class. (c) Which is higher: the average score or the rms score? Why? Figure P18.82 altitude of about 11 km, the temperature is not uniform but decreases with increasing elevation. (a) Show that if the temperature variation is approximated by the linear relationship T = T 0 − α y where T 0 is the temperature at the earth’s surface and T temperature at height y , the pressure p at height y is ln ( p p 0 ) = M g R α ln ( T 0 − α y T 0 ) where P 0 is the pressure at the earth’s surface and M is the molar mass for air. The coefficient α is called the lapse rate of temperature. It varies with atmospheric conditions, but an average value is about 0.6 C°/100 m. (b) Show that the above result reduces to the result of Example 18.4 (Section 18.1) in the limit that α → 0. (c) With α = 0 6 C°/100 m, calculate p for y = 8863 m and compare your answer to the result of Example 18.4. Take T 0 = 288 K and p 0 = 1.00 atm.
DATA The statistical quantities “average value” and “root-mean-square value” can be applied to any distribution. Figure P18.82 shows the scores of a class of 150 students on a 100-point quiz. (a) Find the average score for the class. (b) Find the rms score for the class. (c) Which is higher: the average score or the rms score? Why? Figure P18.82 altitude of about 11 km, the temperature is not uniform but decreases with increasing elevation. (a) Show that if the temperature variation is approximated by the linear relationship T = T 0 − α y where T 0 is the temperature at the earth’s surface and T temperature at height y , the pressure p at height y is ln ( p p 0 ) = M g R α ln ( T 0 − α y T 0 ) where P 0 is the pressure at the earth’s surface and M is the molar mass for air. The coefficient α is called the lapse rate of temperature. It varies with atmospheric conditions, but an average value is about 0.6 C°/100 m. (b) Show that the above result reduces to the result of Example 18.4 (Section 18.1) in the limit that α → 0. (c) With α = 0 6 C°/100 m, calculate p for y = 8863 m and compare your answer to the result of Example 18.4. Take T 0 = 288 K and p 0 = 1.00 atm.
DATA The statistical quantities “average value” and “root-mean-square value” can be applied to any distribution. Figure P18.82 shows the scores of a class of 150 students on a 100-point quiz. (a) Find the average score for the class. (b) Find the rms score for the class. (c) Which is higher: the average score or the rms score? Why?
Figure P18.82
altitude of about 11 km, the temperature is not uniform but decreases with increasing elevation. (a) Show that if the temperature variation is approximated by the linear relationship
T
=
T
0
−
α
y
where T0 is the temperature at the earth’s surface and T temperature at height y, the pressure p at height y is
ln
(
p
p
0
)
=
M
g
R
α
ln
(
T
0
−
α
y
T
0
)
where P0 is the pressure at the earth’s surface and M is the molar mass for air. The coefficient α is called the lapse rate of temperature. It varies with atmospheric conditions, but an average value is about 0.6 C°/100 m. (b) Show that the above result reduces to the result of Example 18.4 (Section 18.1) in the limit that α → 0. (c) With α = 0 6 C°/100 m, calculate p for y = 8863 m and compare your answer to the result of Example 18.4. Take T0 = 288 K and p0 = 1.00 atm.
1. A charge of -25 μC is distributed uniformly throughout a spherical volume of radius 11.5 cm.
Determine the electric field due to this charge at a distance of (a) 2 cm, (b) 4.6 cm, and (c) 25 cm from
the center of the sphere.
(a) =
=
(b) E =
(c)Ẻ =
=
NC NC NC
1.
A long silver rod of radius 3.5 cm has a charge of -3.9
ис
on its surface. Here ŕ is a unit vector
ст
directed perpendicularly away from the axis of the rod as shown in the figure.
(a) Find the electric field at a point 5 cm from the center of the rod (an outside point).
E =
N
C
(b) Find the electric field at a point 1.8 cm from the center of the rod (an inside point)
E=0
Think & Prepare
N
C
1. Is there a symmetry in the charge distribution? What kind of symmetry?
2. The problem gives the charge per unit length 1. How do you figure out the surface charge density σ
from a?
1. Determine the electric flux through each surface whose cross-section is shown below.
55
S₂
-29
S5
SA
S3
+ 9
Enter your answer in terms of q and ε
Φ
(a) s₁
(b) s₂
=
-29
(C) Φ
զ
Ερ
(d) SA
=
(e) $5
(f) Sa
$6
=
II
✓
-29
S6
+39
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