As it is given that Bulk modulus and density related to velocity of sound
by rearranging the equation we can say
now we need to find the SI unit of Bulk modulus here
we can find it by plug in the units of density and speed here
so SI unit will be
SO above is the SI unit of bulk Modulus
To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.
Newton's second law is defined as
Where,
m = mass
a = acceleration
From this equation we can figure the acceleration out, then
From the cinematic equations of motion we know that
Where,
Final velocity
Initial velocity
a = acceleration
x = displacement
There is not Final velocity and the acceleration is equal to the gravity, then
From the equation of motion where acceleration is equal to the velocity in function of time we have
Therefore the time required is 0.0705s
Answer:
A. Stratosphere is said to be stable layer of the atmosphere when cool air sinks and warm air rises.Due to the fact that cool air has tendency to sink ,the air is not going fluctuating up and down in the stratosphere. This means that the air remains stationary or particles remains there for a very long duration.
B. If the lifted index is negative then the parcel temperature is warmer than the actual temperature. In addition, the parcel that is less warm than the surrounding will be less dense and will rise.
C. The water vapor come from different kinds of fronts; gust fronts from existing storms as their downdraft hits the surface, spreads and lifts air in front, upper air disturbances and surface heating by solar radiation making an unbalanced vertical profile .
D. the threshold used by storm chasers to assess if the dew point temperature is high enough to produce large thunderstorms is moisture ,the surface dew point needs to be 55 degrees fahrenheit or greater for a surface based thunderstorm to occur.
E. Wind shear is the change in wind direction or speed with height in an atmosphere.
Explanation:
Answer:
axial stress in the diagonal bar =36,000 psi
Explanation:
Assuming we have to find axial stress
Given:
width of steel bar: 1.25 in.
height of the steel bar: 3 in
Length of the diagonal member = 20ft
modulus of elasticity E= 30,000,000 psi
strain in the diagonal member ε = 0.001200 in/in
Therefore, axial stress in the diagonal bar σ = E×ε
= 30,000,000 psi× 0.001200 in/in =36,000 psi
