GO Hot chocolate effect . Tap a metal spoon inside a mug of water and note the frequency f i you hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value f s because the tiny air bubbles released by the powder change the water’s bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don’t appreciably change the water’s density or volume or the sound’s wavelength. Rather, they change the value of dV/dp— that is, the differential change in volume due to the differential change in the pressure caused by the sound wave in the water. If f s / f i = 0.333, what is the ratio ( dV / dp ) s /( dV / dp ) i ?
GO Hot chocolate effect . Tap a metal spoon inside a mug of water and note the frequency f i you hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value f s because the tiny air bubbles released by the powder change the water’s bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don’t appreciably change the water’s density or volume or the sound’s wavelength. Rather, they change the value of dV/dp— that is, the differential change in volume due to the differential change in the pressure caused by the sound wave in the water. If f s / f i = 0.333, what is the ratio ( dV / dp ) s /( dV / dp ) i ?
GOHot chocolate effect. Tap a metal spoon inside a mug of water and note the frequency fi you hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value fs because the tiny air bubbles released by the powder change the water’s bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don’t appreciably change the water’s density or volume or the sound’s wavelength. Rather, they change the value of dV/dp—that is, the differential change in volume due to the differential change in the pressure caused by the sound wave in the water. If fs/fi = 0.333, what is the ratio (dV/dp)s/(dV/dp)i?
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
Biology: Life on Earth with Physiology (11th Edition)
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