Given Information:
Magnetic field = B = 1×10⁻³ T
Frequency = f = 72.5 Hz
Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m
Required Information:
Maximum Emf = ?
Answer:
Maximum Emf = 20.66×10⁻¹² volts
Explanation:
The maximum emf generated around the perimeter of a cell in a field is given by
Emf = BAωcos(ωt)
Where A is the area, B is the magnetic field and ω is frequency in rad/sec
For maximum emf cos(ωt) = 1
Emf = BAω
Area is given by
A = πr²
A = π(d/2)²
A = π(7.60×10⁻⁶/2)²
A = 45.36×10⁻¹² m²
We know that,
ω = 2πf
ω = 2π(72.5)
ω = 455.53 rad/sec
Finally, the emf is,
Emf = BAω
Emf = 1×10⁻³*45.36×10⁻¹²*455.53
Emf = 20.66×10⁻¹² volts
Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts
Answer:
18 m
Explanation:
Given : vo = 0 m/s ; t = 3 s; a = 4 m/s^2 ; d = ? m ; average velocity = ? m/s ; fonal velocity = ? m/s
solving for the final velocity, v
v = a * t
v = 4 m/s^2 * 3 s
v = 12 m / s
Solving for the average velocity. avg v
avg v = (vo + v) / 2
avg v = (0 m / s + 12 m/s) / 2
avg v = 6 m / s
Solving for the distance traveled after 3 s
d = avg v * t
d = 6 m / s * 3 s
d = 18 meters
In the first 3s the car travels 18 meters.
Answer:
35 m
0.56 m/s west
Explanation:
A) Total distance is the length of the path taken.
30 m + 5 m = 35 m
B) Velocity is displacement over time. Displacement is the difference between the final position and the initial position.
If west is -x, and east is +x, then:
Δx = -30 m + 5 m
Δx = -25 m
v = Δx / t
v = -25 m / 45 s
v = -0.56 m/s
v = 0.56 m/s west
Answer:
Explanation:
For an electromagnetic wave, the relationship between magnetic field amplitude and electric field amplitude is given by
where
E is the amplitude of the electric field
c is the speed of light
B is the amplitude of the magnetic field
For the electromagnetic wave in this problem, we have
E = 10 V/m is the amplitude of the electric field
So if we solve the formula for B, we find the amplitude of the magnetic field:
Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s
