Answer:
the ball experiences the greater momentum change
Explanation:
You have to take into account that momentum change is given by
where vf and vb are the speed of the object after and before the impact.
In the case of the ball you have
where the minus of vf is included due to the motion is in an opposite direction regarding with vb
And for the lump
Hence, the ball experiences the greater change
hope this helps!!
Answer:
<em>The rubber band will be stretched 0.02 m.</em>
<em>The work done in stretching is 0.11 J.</em>
Explanation:
Force 1 = 44 N
extension of rubber band = 0.080 m
Force 2 = 11 N
extension = ?
According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.
F = ke
where k = constant of elasticity
e = extension of the material
F = force applied.
For the first case,
44 = 0.080K
K = 44/0.080 = 550 N/m
For the second situation involving the same rubber band
Force = 11 N
e = 550 N/m
11 = 550e
extension e = 11/550 = <em>0.02 m</em>
<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.
potential energy stored =
==> = <em>0.11 J</em>
Answer: 38 weeks (266 days) from the date of conception.
Explanation: www.momjunction.com/pregnancy-due-date-calculator/
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