Shrinking Loop . A circular loop of flexible iron wire has an initial circumference of 165.0 cm, but its circumference is decreasing at a constant rate of 12.0 cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop and with magnitude 0.500 T. (a) Find the emf induced in the loop at the instant when 9.0 s have passed, (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Shrinking Loop . A circular loop of flexible iron wire has an initial circumference of 165.0 cm, but its circumference is decreasing at a constant rate of 12.0 cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop and with magnitude 0.500 T. (a) Find the emf induced in the loop at the instant when 9.0 s have passed, (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 165.0 cm, but its circumference is decreasing at a constant rate of 12.0 cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop and with magnitude 0.500 T. (a) Find the emf induced in the loop at the instant when 9.0 s have passed, (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Question 6:
Chlorine is widely used to purify municipal water supplies and to treat swimming pool
waters. Suppose that the volume of a particular sample of Cl₂ gas is 8.70 L at 895 torr
and 24°C.
(a) How many grams of Cl₂ are in the sample?
⚫ Atomic mass of CI = 35.453 g/mol
• Molar mass of Cl₂ = 2 x 35.453 = 70.906 g/mol
Solution:
Use the Ideal Gas Law:
Step 1: Convert Given Values
• Pressure: P = 895 torr → atm
PV=
= nRT
1
P = 895 ×
= 1.1789 atm
760
•
Temperature: Convert to Kelvin:
T24273.15 = 297.15 K
• Gas constant: R = 0.0821 L atm/mol. K
Volume: V = 8.70 L
Step 2: Solve for n
.
PV
n =
RT
n =
(1.1789)(8.70)
(0.0821)(297.15)
10.25
n =
= 0.420 mol
24.405
Step 3: Calculate Mass of Cl₂
Final Answer: 29.78 g of Cl₂.
mass nx M
mass=
(0.420)(70.906)
mass=
29.78 g
E1
R₁
w
0.50
20 Ω
12
R₁₂
ww
ΒΩ
R₂
60
E3
C
RA
w
15 Ω
E2
0.25
E4
0.75 Ω
0.5 Ω
Solve plz
Chapter 29 Solutions
University Physics with Modern Physics, Volume 1 (Chs. 1-20) (14th Edition)
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