Answer:
The observer hears a loud sound
Explanation:
In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.
You first calculate the distance between the observer and the loudspeakers.
The distances are given by:
d1: distance to loudspeaker A = 2.10m
d2: distance to loudspeaker B
Next, you calculate the wavelength of the sound waves by using the following formula:
vs: speed of sound = 343 m/s
f: frequency of the waves = 400Hz
λ: wavelength
Next, you calculate the path difference between the distance from the observer to the loudspeakers:
You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:
Then, there will be a constructive interference, and the sound who the observer hears is loud.
RC circuit determines the capacitor's charging rate.
- In RC (resistive and capacitive) circuits, a capacitor's time constant is the number of seconds required to charge it to 63.2% of the input voltage.
- This duration is described by a single time constant. After two time constants, the capacitor will be charged to 86.5% of the input voltage.
- The RC time constant, also referred to as tau, is the time constant (in seconds) of an RC circuit and is obtained by multiplying the circuit resistance (in ohms) by the circuit capacitance (in farads), This transient reaction time T is stated in terms of = R x C, where R is the resistor value in ohms and C is the capacitor value in farads.
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Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:
0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:
Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:
5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:
10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:
10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]
Answer:
The answer is True
Explanation:
Statistical Multiplexing is considered an example of communication link sharing which makes it comparable to DBA (Dynamic Bandwidth Allocation). Here, communication channels are broken down into data streams to optimize the communication process.
In Statistical Time-division Multiplexing, time slots are allocated to data streams for communication optimization. This method makes sure that no time slot or bandwidth is wasted.
Hence, the sum of combined circuits must not be equal to the capacity of the circuit to work effectively.
