Vf^2 = Vi^2 + 2ad
Vf^ = 0 + 2(-9.8)(-12)
Vf^2 = 235.2
Vf = 15.3 m/s
The correct answer is b) 15 m/s
Answer:
w_f = m*V*cos(Q_n) / L*(m+M)
Explanation:
Given:
- mass of the putty ball m
- mass of the rod M
- Velocity of the ball V
- Length of the rod L
- Angle the ball makes before colliding with rod Q_n
Find:
What is the angular speed ωf of the system immediately after the collision,
Solution:
- We can either use conservation of angular momentum or conservation of Energy. We will use Conservation of angular momentum of a system:
L_before = L_after
- Initially the rod is at rest, and ball is moving with the velocity V at angle Q from normal to the rod. We know that the component normal to the rod causes angular momentum. Hence,
L_before = L_ball = m*L*V*cos(Q_n)
- After colliding the ball sicks to the rod and both move together with angular speed w_f
L_after = (m+M)*L*v_f
Where, v_f = L*w_f
L_after = (m+M)*L^2 * w_f
- Now equate the two expression as per conservation of angular momentum:
m*L*V*cos(Q_n) = (m+M)*L^2 * w_f
w_f = m*V*cos(Q_n) / L*(m+M)
<h2>For Second Solid Lumped System is Applicabe</h2>
Explanation:
Considering heat transfer between two identical hot solid bodies and their environments -
- If the first solid is dropped in a large container filled with water, while the second one is allowed to cool naturally in the air than for second solid, the lumped system analysis more likely to be applicable
- The reason is that a lumped system analysis is more likely to be applicable in the air than in water as the convection heat transfer coefficient so that the Biot number is less than or equal to 0.1 that is much smaller in air
Biot number = the ratio of conduction resistance within the body to convection resistance at the surface of the body
∴ For a lumped system analysis Biot number should be less than 0.1
Answer:
(a). 12 plants
(b). 3171 $
Explanation:
(a)first convert units of 100 billion kWh/year into Watts(W)
also convert the units of 1000 MW into Watts(W)
1 billion = 10^9
1 year = 365*24 = 8760 hrs
so
100 billion kWh/year = 1
= 
W
1000 MW = 
no. of plants = 
= 11.4
So 12 plants required
(b)
savings = unit price*total units
= 
= 3170.9 =3171 $
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
