Answer:
Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.
Explanation:
Given:
Speed of sound in air = 320 m/s
Speed of sound in water = 1600 m/s
Time taken to reach certain distance in air = 2.5 sec
a.
We have to find the distance traveled by sound in air.
Distance = Product of speed and time.
⇒ 
⇒ 
⇒ 
meters.
b.
Now we have to find how much time the sound will take to travel in water.
⇒ Time = Ratio of distance and speed
⇒ 
⇒ 
<em> ...distance = 800 m and speed = 1600 m/s</em>
⇒ 
⇒ 
seconds.
Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.
Answer:
Vertical distance= 3.3803ft
Explanation:
First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:
Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h
Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h
time= 0.00012731h × (3600s/h)= 0.458316s
With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:
Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m
Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft
This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.
The correct answer is Option C) Sample C would be best, because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest.
As the coefficient of absorption would define the energy present in the reflected wave, the material C has the highest percentage of absorption i.e. 62% and would be best suitable to make a sound proof room.
Answer:
<h2>
The potential difference increases </h2>
Explanation:
from the relation 
where E= electric field (force per coulomb)
V= voltage
d= distance
Hence the voltage is going to be V= E×d.
Therefore this means that increasing the distance increases the voltage.
Answer:
a) L = 0.75m f₁ = 113.33 Hz
, f₃ = 340 Hz, b) L=1.50m f₁ = 56.67 Hz
, f₃ = 170 Hz
Explanation:
This resonant system can be simulated by a system with a closed end, the tile wall and an open end where it is being sung
In this configuration we have a node at the closed end and a belly at the open end whereby the wavelength
With 1 node λ₁ = 4 L
With 2 nodes λ₂ = 4L / 3
With 3 nodes λ₃ = 4L / 5
The general term would be λ_n= 4L / n n = 1, 3, 5, ((2n + 1)
The speed of sound is
v = λ f
f = v / λ
f = v n / 4L
Let's consider each length independently
L = 0.75 m
f₁ = 340 1/4 0.75 = 113.33 n
f₁ = 113.33 Hz
f₃ = 113.33 3
f₃ = 340 Hz
L = 1.5 m
f₁ = 340 n / 4 1.5 = 56.67 n
f₁ = 56.67 Hz
f₃ = 56.67 3
f₃ = 170 Hz
