The weight changes but the mass will stay the same.
Answer:
660 J/kg/°C
Explanation:
Heat lost by metal = heat gained by water
-m₁C₁ΔT₁ = m₂C₂ΔT₂
-(0.45 kg) C₁ (21°C − 80°C) = (0.70 kg) (4200 J/kg/°C) (21°C − 15°C)
C₁ = 660 J/kg/°C
Answer:
f1 = 58.3Hz, f2 = 175Hz, f3 = 291.6Hz
Explanation:
lets assume speed of sound is 350 m/s.
frequencies of a standing wave modes of an open-close tube of length L
fm = m(v/4L)
where m is 1,3,5,7......
and fm = mf1
where f1 = fundamental frequency
so therefore: f1 = 350 x 4 / 1.5
f1 = 58.3Hz
f2 = 3 x 58.3
f2 = 175Hz
f3 = 5 x 58.3
f3 = 291.6Hz
Answer:
h = 1.02 m
Explanation:
This is a fluid mechanics exercise, where the pressure is given by
P =
+ ρ g h
The gauge pressure is
P -
= ρ g h
In this case the upper part of the tube we have the atmospheric pressure. and the diver can exert a pressure 10 KPa below the outside pressure, this must be the gauge pressure
= P - 
= ρ g h
h =
/ ρ g
calculate
h = 10 103 / (1000 9.8)
h = 1.02 m
This is the depth at which man can breathe
Mercury has a high boiling point of 357 degrees C.
Mercury has a freezing point of −39 degrees C.