The decibel system of sound intensity operates by a logarithmic scale, meaning that sound intensity increases exponentially in relation to the decibel rating.
For decibels, the equation between intensity and the dB equivalent is:
dB = 10log(i),
where “i” is the intensity of the sound. The ten in front of the log means that an increase in ten dB results in a tenfold increase in sound intensity; for example, a 30 dB sound is ten times softer than a 40 dB sound.
In this case, a sound with a dB of 80 would be 1000 times more intense than a 50 dB sound, so the decibel rating of B is 80.
Hope this helps!
To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.
Ff = 0.5 * 16 * 9.8 = 78.4 N
a = 4.9 m/s^2
If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?
In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.
Ff = 0.28 * 16 * 9.8 = 43.904 N
Net force = 78.4 – 43.904 = 34.496 N
To determine the acceleration, divide by the mass of the crate.
a = 34.496 ÷ 16 = 2.156 m/s^2
M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h = 0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
W = 1173.5KW
Let say the point is inside the cylinder
then as per Gauss' law we have
here q = charge inside the gaussian surface.
Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.
we will calculate the charge first which is given as
now using the equation of Gauss law we will have
now we will have
Now if we have a situation that the point lies outside the cylinder
we will calculate the charge first which is given as it is now the total charge of the cylinder
now using the equation of Gauss law we will have
now we will have
Answer:
a) Acceleration of the car is given as
b) Acceleration of the truck is given as
Explanation:
As we know that there is no external force in the direction of motion of truck and car
So here we can say that the momentum of the system before and after collision must be conserved
So here we will have
now we have
a) For acceleration of car we know that it is rate of change in velocity of car
so we have
b) For acceleration of truck we will find the rate of change in velocity of the truck
so we have