Classius claperyon equation
In (P2/ P2) = ΔHvap/R) × (1/T2-1/T1)
T2 occurs at normal boiling when vapor pressure P2 = 1 atm.
P1 = 55.1 mmHg, P2 = 1 atm = 760mmHg
T1 = 35°c = 308.15k, T2 =
ΔHvap = 32.1kJ/mol = 32100 J/mol
In (760/55.1) = (-32100/ 8.314) × ( 1/T2 - 1/308.15)
The normal boiling point T2 = 390k = 117°c
Answer:
lambda = 343 m/s divided by 340 Hz = 1.009 seconds
Hope it helps and have a wonderful day!
I think it's D because sunlight is a solar system
Answer:
Explanation:
spring constant k = 425 N/m
a ) At the point of equilibrium
restoring force = frictional force
= kx = 10 N
425 x = 10
x = 2.35 cm
b )
Work done by frictional force
= -10 x 2.35 x 10⁻² x 2 J ( Distance is twice of 2.35 cm )
= - 0.47 J
= Kinetic energy remaining with the cookie as it slides back through the position where the spring is unstretched .
= 425 - 0.47
= 424.53 J
=
