The sound wave will have traveled 2565 m farther in water than in air.
Answer:
Explanation:
It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.
Distance = Velocity × Time.
So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.
As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.
Distance = 343 × 2.25 =771.75 m
And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.
Distance = 1483×2.25=3337 m.
Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.
Difference in distance covered in water and air = 3337-772 m = 2565 m
So the sound wave will have traveled 2565 m farther in water than in air.
Answer:
Minimum elastic modulus of fiber = 455.64 GPa
Explanation:
Contents of composite material = Epoxy and Unidirectional fibers
Elastic modulus of epoxy = 3.5 GPa
Elastic modulus of composite material = 320 GPa
Volume fraction of fiber = 70 %
Volume fraction of epoxy = 100 - 70 = 30%
Elastic modulus of composite material = 3.5 x 0.3 + Elastic modulus of fiber x 0.7 = 320
0.7 x Elastic modulus of fiber = 320 - 1.05 = 318.95
Elastic modulus of fiber = 455.64 GPa
Minimum elastic modulus of fiber = 455.64 GPa
Given: Heat Qout means useful work = 2800 J
Heat Qin = 8900 J
Required; Efficiency = ?
Formula: Efficiency = Qout/Qin = x 100%
= 2800 J/8900 J = 0.3146 X 100 %
Efficiency = 31.46%
Answer:
it tells you that the speed increases until about 20 seconds then keeps a steady pace for 20 seconds then the speed drops and stops at 55 seconds in the process.
Air that is not saturated will cool or heat at a rate of 10 degrees C/1000 meters as it rises or descends, respectively
The rate of the change in temperature which is observed while moving upward in the earth's atmosphere with elevation. It can be positive, negative and zero when the temperature decreases, increases or is constant with the elevation respectively.
For the atmosphere, the drop in temperature of rising air that is unsaturated air is about 10 degrees C/1000 meters (5.5 °F per 1000 feet) altitude.
That means if a there occur a rise of 1000 m , then the temperature of that thing will decrease to 10 degrees. Every 10°C of temperature from the given temperature will decrease at every rise of 1000 m .Air that is not saturated will cool or heat at a rate of 10 degrees C/1000 meters as it rises or descends, respectively
To learn more about unsaturated air here
brainly.com/question/13434971?referrer=searchResults
#SPJ4
