Answer:
a. B = 0.20T
b. u = 17230.6 J/m³
c. E = 0.236J
d. L = 5.84*10^-5 H
Explanation:
a. In order to calculate the magnetic field in the solenoid you use the following formula:
(1)
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
n: turns of the solenoid = 460
L: length of the solenoid = 25.0cm = 0.25m
i: current = 90.0A
You replace the values of the parameters in the equation (1):
The magnetic field in the solenoid is 0.20T
b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:
The energy density is 17230.6 J/m³
c. The total energy contained in the solenoid is:
(2)
V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:
A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2
Then, the energy contained in the solenoid is:
The energy contained is 0.236J
d. The inductance of the solenoid is calculated as follow:
The inductance of the solenoid is 5.84*10^-5 H
Answer:
0.36 A.
Explanation:
We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:
Resistor 1 (R₁) = 35 Ω
Resistor 2 (R₂) = 20 Ω
Equivalent Resistance (Rₑq) =?
Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:
Rₑq = (R₁ × R₂) / (R₁ + R₂)
Rₑq = (35 × 20) / (35 + 20)
Rₑq = 700 / 55
Rₑq = 12.73 Ω
Next, we shall determine the total resistance in the circuit. This can be obtained as follow:
Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω
Resistor 3 (R₃) = 15 Ω
Total resistance (R) in the circuit =?
R = Rₑq + R₃ (they are in series connection)
R = 12.73 + 15
R = 27.73 Ω
Finally, we shall determine the current. This can be obtained as follow:
Total resistance (R) = 27.73 Ω
Voltage (V) = 10 V
Current (I) =?
V = IR
10 = I × 27.73
Divide both side by 27.73
I = 10 / 27.73
I = 0.36 A
Therefore, the current is 0.36 A.
Answer:
1/4 times your earth's weight
Explanation:
assuming the Mass of earth = M
Radius of earth = R
∴ the mass of the planet= 4M
the radius of the planet = 4R
gravitational force of earth is given as = 
where G is the gravitational constant
Gravitational force of the planet = 
=
=
recall, gravitational force of earth is given as =
∴Gravitational force of planet = 1/4 times the gravitational force of the earth
you would weigh 1/4 times your earth's weight
Answer
given,
weight of the oak board = 600 N
Weight of Joe = 844 N
length of board = 4 m
Joe is standing at 1 m from left side
vertical wire is supporting at the end.
Assuming the system is in equilibrium
T₁ and T₂ be the tension at the ends of the wire
equating all the vertical force
T₁ + T₂ = 600 + 844
T₁ + T₂ = 1444...........(1)
taking moment about T₂
T₁ x 4 - 844 x 3 - 600 x 2 = 0
T₁ x 4 = 3732
T₁ = 933 N
from equation (1)
T₂ = 1444 - 933
T₂ = 511 N
a)
consider the motion in upward direction as positive and down direction as negative
Y₀ = initial position of the stone = 20 m
v₀ = initial velocity of the stone = 10 m/s
a = acceleration = - 9.8 m/s²
Y = final position of the stone when it reach the maximum height
v = final velocity at the maximum height = 0 m/s
t = time taken to reach the maximum height
Using the equation
v² = v₀² + 2 a (Y - Y₀)
0² = 10² + 2 (- 9.8) (Y - 20)
Y = 25.1 m
also using the equation
v = v₀ + a t
inserting the values
0 = 10 + (- 9.8) t
t = 1.02 sec
b)
consider the motion in upward direction as positive and down direction as negative
Y₀ = initial position of the stone = 20 m
v₀ = initial velocity of the stone = 10 m/s
a = acceleration = - 9.8 m/s²
Y = final position of the stone when it reach the ground = 0 m
t = time taken to reach the ground
Using the equation
Y = Y₀ + v₀ t + (0.5) a t²
0 = 20 + 10 t + (0.5) (- 9.8) t²
t = 3.3 sec
