Answer: a) W(earth) = 935.62 lbs
b) Mass of rocks in slugs = 29.06 slugs
Explanation:
a) From Newton's law, W = mg. Whether on the moon or on earth. Although, the mass of the rocks everywhere is the same, that is, mass of rocks on the moon = mass of rocks on earth.
W(moon) = mg(moon)
W(moon) = 154 lbs
g(moon) = 5.30 ft/s2
m = W(moon)/g(moon) = 154/5.3 = 29.06 lb.s2/ft
W(earth) = m g(earth)
g(earth) = 32.2 ft/s2
W(earth) = 29.06 × 32.2 = 935.62 lbs.
b) A slug = 1 lb.s2/ft, therefore the mass of the rocks in slugs is 29.06 slugs.
QED!
Answer:
for a) F= 744.97 N
for a) F= 167.85 N
for a) F= 764.57 N
Explanation:
the pressure developed by the piston should be higher than the saturated vapor pressure of water for boiling point at T=120 to ensure boiling.
Then from steam tables
T= 120°C → P required=Pr= 198.67 kPa
then the pressure developed by the piston is
P = (m*g + F)/A
where m= mass of the piston ,g= gravity F= force required and A= area of the piston
then
Pr = P = (m*g + F)/A
F = Pr*A-m*g
since A= π/4*D²
F =π/4* Pr*D²-m*g
replacing values
F =π/4* Pr*D²-m*g = π/4*198.67 *10³Pa*(0.07m)² -2kg* 9.8m/s²
F= 744.97 N
b) for T₂=80°C → Pr₂=47.41 kPa
F₂ =π/4* Pr₂*D²-m*g = π/4*47.41*10³Pa*(0.07m)² -2kg* 9.8m/s²
F₂= 167.85 N
c) for m=0 (mass of the piston neglected) ,the force required is
F₃ =π/4*Pr*D² = π/4*198.67 *10³Pa*(0.07m)²= 764.57 N
F₃ =764.57 N
Answer:
what we have to answer please mention questions
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- <em>I</em><em> </em><em>feel</em><em> </em><em>happy</em><em> </em><em>when</em><em> </em><em>I</em><em> </em><em>listen</em><em> </em><em>to</em><em> </em><em>this</em><em> </em><em>song</em><em> </em><em>because</em><em> </em><em>it</em><em> </em><em>shows</em><em> </em><em>the</em><em> </em><em>connection</em><em> </em><em>between</em><em> </em><em>a</em><em> </em><em>mother</em><em> </em><em>and</em><em> </em><em>daughter</em><em>.</em>
- <em>Also</em><em>,</em><em> </em><em>The</em><em> </em><em>song</em><em> </em><em>compares</em><em> </em><em>Sunshine</em><em> </em><em>with</em><em> </em><em>Happiness</em><em>.</em>
Answer:
a) 1 / 3 N
b) 5/3 m/s
Explanation:
For constant speed V at the interface of the two fluids the net force is zero.
Hence, F top = F bottom thus the shear stresses at top and bottom must be the same.
For force of top plate