if a mass on a spring bobs up nad down completing 2 full cycles every section. WHat are the periouds and frequency of the mass
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Answer:
Given that
Dry-bulb temperature(T) =24°C
Wet-bulb temperature(Tw) = 17°C
Pressure ,P = 1 atm
As we know that psychrometric chart are drawn at constant pressure.
From the diagram
ω= specific humidity
Lets take these two lines Dry-bulb temperature(T) line and Wet-bulb temperature(Tw) cut at point P
From chart at point P
a)
Specific humidity,ω = 0.00922 kg/kg
b)
The enthalpy ( h)
h=47.59 KJ/kg
c)
The relative humidity, RH
RH= 49.58 %
d)
Specific volume ,
v= 0.853 m³/kg
1 kilometre=1000 metre
1 hour = 3600 second
The initial velocity of car A is 35.0 km/hr i.e
= 9.72 m/s
The initial velocity of car B is 45 km/hr =12.5 m/s
The initial velocity of car C is 32 km/hr = 8.89 m/s
The initial velocity of car D is 110 km/hr=30.56 m/s
The acceleration of car A is given as
The time taken by car A = 15 min.
From equation of kinematics we know that-
v= u+at [Here v is the final velocity and a is the acceleration and t is the time]
Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s
=11.456111106 m/s
The acceleration of B is given as
The time taken by car B =20 min
The final velocity of B is -
v= u+at
= u-at [Here a is negative due to deceleration]
=12.5 m/s +[0.0011574074074×20×60]
=13.8888888.....
=13.9
The acceleration of C is given as
The time taken by car C =30 min
The final velocity of C is-
v = u+at
=8.89 m/s+[0.003086419753×30×60] m/s
=14.4455555555..m/s
=14.45 m/s
The car C is decelerating.The deceleration is given as-
The time taken by car D= 45 min.
The final velocity of the car D is-
v =u+at
=30.56 -[0.00462962962962×45×60]m/s
=18.06 m/s
Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.