Answer:
The reactance of the capacitor
Explanation:
In an AC circuit containing different elements (capacitors, resistors and inductors), we cannot simply calculate the equivalent resistance of the circuit, so another quantity is used, which is called reactance.
For a capacitor, the reactance is given by:
where:
f is the frequency of the AC current in the circuit
C is the capacitance of the capacitor
The reactance has a similar meaning to that of the resistance for a DC current. In fact, we notice that:
- When f=0 (which means we are in regime of DC current, because the current never changes direction), the reactance is infinite. This is correct: in a DC circuit, the capacitor does not let current pass through it, so it like it has infinite resistance (=infinite reactance)
- When f tends to infinite, the reactance becomes zero: in such situation, the current in the circuit changes direction so quickly that the capacitor has no enough time to "block" the current in the circuit, so it like it has almost zero resistance (zero reactance).
Answer:
1456 N
Explanation:
Given that
Frequency of the piano, f = 27.5 Hz
Entire length of the string, l = 2 m
Mass of the piano, m = 400 g
Length of the vibrating section of the string, L = 1.9 m
Tension needed, T = ?
The formula for the tension is represented as
T = 4mL²f²/ l, where
T = tension
m = mass
L = length of vibrating part
F = frequency
l = length of the whole part
If we substitute and apply the values we have Fri. The question, we would have
T = (4 * 0.4 * 1.9² * 27.5²) / 2
T = 4368.1 / 2
T = 1456 N
Thus, we could conclude that the tension needed to tune the string properly is 1456 N
I don't know about it your answer will give another people
Answer:
Explanation:
From the second law of Newton movement laws, we have:
, and we know that a is the acceleration, which definition is:
, so:
The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):
(This is the indefinite integral), the definite one is:
The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.
