Answer:
a) 5.63 atm
Explanation:
We can use combined gas law
<em>The combined gas law</em> combines the three gas laws:
- Boyle's Law, (P₁V₁ =P₂V₂)
- Charles' Law (V₁/T₁ =V₂/T₂)
- Gay-Lussac's Law. (P₁/T₁ =P₂/T₂)
It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.
P₁V₁/T₁ =P₂V₂/T₂
where P = Pressure, T = Absolute temperature, V = Volume occupied
The volume of the system remains constant,
So, P₁/T₁ =P₂/T₂
a) 
Answer:
9.25 x 10^-4 Nm
Explanation:
number of turns, N = 8
major axis = 40 cm
semi major axis, a = 20 cm = 0.2 m
minor axis = 30 cm
semi minor axis, b = 15 cm = 0.15 m
current, i = 6.2 A
Magnetic field, B = 1.98 x 10^-4 T
Angle between the normal and the magnetic field is 90°.
Torque is given by
τ = N i A B SinФ
Where, A be the area of the coil.
Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²
τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°
τ = 9.25 x 10^-4 Nm
thus, the torque is 9.25 x 10^-4 Nm.
Answer:
It would take
time for the capacitor to discharge from
to
.
It would take
time for the capacitor to discharge from
to
.
Note that
, and that
.
Explanation:
In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is
, and the initial charge of the capacitor be
. Then at time
, the charge stored in the capacitor would be:
.
<h3>a)</h3>
.
Apply the equation
:
.
The goal is to solve for
in terms of
. Rearrange the equation:
.
Take the natural logarithm of both sides:
.
.
.
<h3>b)</h3>
.
Apply the equation
:
.
The goal is to solve for
in terms of
. Rearrange the equation:
.
Take the natural logarithm of both sides:
.
.
.
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A