Is the equation 344+0.6x(t-20)? I'm not getting this at all?
How long does it take for sound to arrive at the back of a big auditorium if the temperature is 33 oC and the stage is 29 m away? answer in seconds
The Eq I have is:
V = 331.4 + 0.6T.
T = Temp. in deg celsius.
V = Velocity in m/s.
V = 331.4 + 0.6*33,
V = 331.4 + 19.8 = 351.2m/s.
t = d/V = 29m / 351.2m/s = 0.08257s.
post
To calculate the time it takes for sound to arrive at the back of a big auditorium, we can use the formula:
Distance = Speed × Time
The speed of sound in air changes with temperature. For temperatures close to room temperature, the speed of sound can be approximated by the formula:
Speed = 331.4 + 0.6 × Temperature
Given that the temperature is 33°C, the speed of sound can be calculated as:
Speed = 331.4 + 0.6 × 33
= 331.4 + 19.8
= 351.2 meters per second (m/s)
Now, we can use the speed of sound to calculate the time it takes for sound to travel a distance of 29 meters.
29 = 351.2 × Time
Dividing both sides of the equation by 351.2, we have:
Time = 29 / 351.2
≈ 0.0826 seconds (rounded to four decimal places)
Therefore, it takes approximately 0.0826 seconds for sound to arrive at the back of the auditorium.
To determine the time it takes for sound to arrive at the back of a big auditorium, we need to use the formula for the speed of sound in air: v = 331.5 + 0.6T, where v is the speed of sound in meters per second (m/s) and T is the temperature in degrees Celsius (°C).
Given that the temperature is 33 °C, we can substitute that value into the formula: v = 331.5 + 0.6(33) = 331.5 + 19.8 = 351.3 m/s.
To calculate the time it takes for sound to travel from the stage to the back of the auditorium, we'll use the formula: time = distance ÷ speed.
Given that the distance from the stage to the back of the auditorium is 29 meters, we can substitute that value into the formula: time = 29 ÷ 351.3 = 0.0827 seconds.
Therefore, it takes approximately 0.0827 seconds for sound to arrive at the back of the big auditorium.