Explanation:
The given data is as follows.
Length of beam, (L) = 5.50 m
Weight of the beam, (
) = 332 N
Weight of the Suki, (
) = 505 N
After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

= 0
x = 
= 
= 0.986 m
Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.
Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.
Answer:
18,850 Hz
Explanation:
We need to figure out the wavelength of the sound wave.
Thus,
Wavelength = 1000 * Lowest Amplitude Wave
Wavelength = 1000 * 2.0 * 10^(-5)
Wavelength = 0.02
Or,

Now, we need the frequency of this wave. It goes by the formula:

Where
f is the frequency in Hz
v is the speed of sound in air (to be 377 m/s)
is the wavelength (we found to be 0.02)
Substituting, we find the frequency:

The wave has frequency of 18,850 Hz
The answer should be the letter C...
Answer:
B. closed series circuit because the resistors are one after the other and there is no open switch.
Explanation:
Answer:
w = √ 1 / CL
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
Explanation:
This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.
In these circuits the impedance is
X = √ (R² + (
-
)² )
where Xc and XL is the capacitive and inductive impedance, respectively
X_{C} = 1 / wC
X_{L} = wL
From this expression we can see that for the resonance frequency
X_{C} = X_{L}
the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
V = IR
Since the contribution of the two other components is canceled, this occurs for
X_{C} = X_{L}
1 / wC = w L
w = √ 1 / CL