Answer:
The water particles move perpendicular to the source of the sound wave.
Explanation:
Water particles are transverse waves
1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum
The force used to kick the ball is 1000N
The mass of the ball is 0.8 kg
Time is 0.8 seconds
Therefore the velocity can be calculated as follows
F= Mv-mu/t
1000= 0.8(v) - 0.8(0)/0.8
1000= 0.8v- 0.8/0.8
Cross multiply both sides
1000(0.8) = 0.8v
800= 0.8v
divide both sides by the coefficient of v which is 8
800/0.8= 0.8v/0.8
v= 1000m/s
Hence the velocity is 1000m/s
Answer:
7.06MN
Explanation:
length = 60m
Width = 10m
Height = 12 m
Let Fb = Force buoyant
Fb = pgV = mg
p(density) = rho
The density of water in this case = 1000
g = 9.8 m/s
Volume = lenght*width*height
= 60*10*12
= 7200m^3
So we have
(1000 * 9.8 * 7200) = mg
= 70560000
7.06 MN
Answer:
a) T2 = 133.5°C
x2 = 0.364
b) Qout = 3959.6 kJ
Explanation:
For this exercise we will make two assumptions: the tank is stationary, the kinetic and potential energy are equal to zero. The second assumption is that there are interactions between the two tanks. The contents of both tanks will be a closed system. The energy balance equals:
ΔEsystem = Ein - Eout
-Qout = ΔUA + ΔUB = (m*(u2-u1))A + (m*(u2-u1))B
The steam properties for the two tanks in the initial state can be found in Table A-4 to table A-6 of Cengel:
P1A = 1000 kPa
T1A = 300°C
v1A = 0.25799 m^3/kg
u1A = 2793.7 kJ/kg
T1B = 150°C
x1 = 0.5
vf = 0.001091 m^3/kg
uf = 631.66 kJ/kg
vg = 0.39248 m^3/kg
ufg = 1927.4 kJ/kg
v1B = vf + x1*vfg = 0.001091 + (0.5*(0.39248-0.001091)) = 0.19679 m^3/kg
u1B = uf + x1*ufg = 631.66 + (0.5*1927.4) = 1595.4 kJ/kg
V = VA + VB = mA*v1A + mB*v1B = (2*0.25799) + (3*0.19679) = 1.106 m^3
m = mA + mB = 3+2 = 5 kg
the specific volume will be equal to:
v2 = V/m = 1.106/5 = 0.2213 m^3/kg
With these calculations, we can looking the new properties in the same tables:
P2 = 300 kPa
v2 = 0.2213 m^3/kg
T2 = Tsat, 300 kPa = 133.5°C
x2 =(v2-vf)/(vg-vf) = (0.22127-0.001073)/(0.60582-0.001073) = 0.364
u2 =uf + x2*ufg = 561.11 + (0.364*1982.1) = 1282.8 kJ/kg
-Qout = (2*(1282.8-2793.7)) + (3*(1282.8-1595.4)) = -3959.6 kJ
Qout = 3959.6 kJ