Hello! Please answer my question that I just posted I really need answers. I don’t know sorry hope u fine someone that does know
Answer:
The total length of wire is 0.24 m.
Explanation:
Number of turns, N = 270
magnetic field, B = 0.48 T
frequency, f = 60 Hz
rms value of emf = 120 V
maximum value of emf, Vo = 1.414 x 120 = 169.68 V
let the area of square is A and the side is L.
The maximum emf is given by
Vo = N B A w
169.68 = 270 x 0.48 x A x 2 x 3.14 x 60
A = 3.5 x 10^-3 m^2
So,
L = 0.0589 m
Total length of wire, P = 4 L = 4 x 0.0589 = 0.24 m
Hello!
I believe C) immunizations are a high-priority health in the United States. First let's talk about what immunizations are. Immunizations are a liquid that is injected into your body by needle. This liquid then enters your blood stream and it gets pumped into all of your body because of your circulatory system. So it spreads all around your body and when you get sick, that vaccine helps to get rid of the disease way quicker or even not at all. The U.S government are in high demand to prevent serious disease by giving people the vaccines because many people are getting sick from, for instance, cancer. Although the actually vaccine cannot cure cancer it can prevent it from happening.
I hope it helps!
I hope it helps!
Answer:
I will answer in English.
Here we will use the relation
Velocity*time = distance
So:
a) velocity = 3m/s
time = 2s
Distance = 3m/s*2s = 6m
b) velocity = 2m/s
time = 3.5s
Distance = 2m/s*3.5s = 7m
c) velocity = 10m/s
time = 0.5s
Distance = 10m/s*0.5s = 5m
d) velocity = 4m/s
time = 2.5s
Distance = 4m/s*2.5s = 9m
e) velocity = 1.5m/s
time = 5s
Distance = 1.5m/s*5s = 7.5m
Answer:
Explanation:
Given an RL circuit
A voltage source of.
V = 108V
A resistor of resistance
R = 1.1-kΩ = 1100 Ω
And inductor of inductance
L = 34 H
After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor
A. Time the inductor current will reduce to 12% of it's initial current
Let the initial charge current be Io
Then, final current is
I = 12% of Io
I = 0.12Io
I / Io = 0.12
The current in an inductor RL circuit is given as
I = Io ( 1—exp(-t/τ)
Where τ is time constant and it is given as
τ = L/R = 34/1100 = 0.03091A
So,
I = Io ( 1—exp(-t/τ))
I / Io = ( 1—exp(-t/τ))
Where I/Io = 0.12
0.12 = 1—exp(-t/τ)
0.12 — 1 = —exp(-t/τ)
-0.88 = -exp(-t/0.03091)
0.88 = exp(-t/0.03091)
Take In of both sides
In(0.88) = In(exp(-t/0.03091)
-0.12783 = -t/0.030901
t = -0.12783 × 0.030901
t = 3.95 × 10^-3 seconds
t = 3.95 ms
B. Energy stored in inductor is given as
U = ½Li²
So, the current at this time t = 3.95ms
I = Io ( 1—exp(-t/τ))
Where Io = V/R
Io = 108/1100 = 0.0982 A
Now,
I = Io ( 1—exp(-t/τ))
I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))
I = 0.0982(1—exp(-0.12783)
I = 0.0982 × 0.12
I = 0.01178
I = 11.78mA
Therefore,
U = ½Li²
U = ½ × 34 × 0.01178²
U = 2.36 × 10^-3 J
U = 2.36 mJ
