Combining conservation laws. A 5.00 kg chunk of ice is sliding at 12.0 m/s on the floor of an ice-covered valley when it collides with and sticks to another 5.00 kg chunk of ice that is initially at rest. (See Figure 8.37 ) Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down. How fast are they moving when they reach the valley floor again? ( Hint: Break this problem into two parts—the collision and the behavior after the collision—and apply the appropriate conservation law to each part.) Figure 8.37 Problem 18.
Combining conservation laws. A 5.00 kg chunk of ice is sliding at 12.0 m/s on the floor of an ice-covered valley when it collides with and sticks to another 5.00 kg chunk of ice that is initially at rest. (See Figure 8.37 ) Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down. How fast are they moving when they reach the valley floor again? ( Hint: Break this problem into two parts—the collision and the behavior after the collision—and apply the appropriate conservation law to each part.) Figure 8.37 Problem 18.
Combining conservation laws. A 5.00 kg chunk of ice is sliding at 12.0 m/s on the floor of an ice-covered valley when it collides with and sticks to another 5.00 kg chunk of ice that is initially at rest. (See Figure 8.37) Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down. How fast are they moving when they reach the valley floor again? (Hint: Break this problem into two parts—the collision and the behavior after the collision—and apply the appropriate conservation law to each part.)
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
Chapter 8 Solutions
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