24 miles per second
56/9=6
6+6+6+6=24
Answer:
Assume two identical cans filled with two types of soup having same mass are rolling down on an inclined plane in same conditions. In terms of inertia different types of soup will indicate different viscosity. The higher viscosity fillings indicates more part of the soup mass is rotating together with the can’s body. This means that for the can with lower viscosity soup has a lower moment of inertia and the can with higher viscosity has higher moment of inertia while the same gravity makes them to roll.
incline angle = θ ; can's mass = m ; Radius of the can's = R , Angular acceleration for Can 1 = α1 ; Angular acceleration for Can 2 = α2
T1 = Inertia of Can with high viscosity soup
T2 = Inertia of Can with low viscosity soup
M1 rolling moment of Can 1
M2 rolling moment of Can 2
equation is given by
T1*α1 = M1 - (a)
T2*α2 = M2 - (b)
M1 = M2 = m*g*R*sin(θ). (c)
as assumed T1 > T2
from the three equation (a), (b) & (c)
the α2 > α1
Angular acceleration of Can 2 is higher than Can 1. Already stated that Can 1 has more viscous soup as compared to Can 2.
Answer: 25N
method: total force in the right hand direction is 100N and total force in the left hand direction is 125N. To get the net force, we add forces if they are in the same direction and substract if they are in opposite directions. since 100N and 125N are in opposite directions, we substract the larger value from the smaller value. Then we get 25N in the left hand direction as the final answer.
Answer:
ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)
Explanation:
Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :
ΔS al ≥ ∫dQ/T
if the heat transfer is carried out reversibly
ΔS al =∫dQ/T
in the surroundings
ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K
the total entropy change will be
ΔS total = ΔS al + ΔS surr
ΔS total ≥ ΔS al + (-ΔS al) =
ΔS total ≥ 0
the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings
<span> </span>For any prism-shaped geometry, the volume
(V) is assumed by the product of cross-sectional area (A) and height (h).
<span> V = Ah </span>
<span>
Distinguishing with respect to time gives the
relationship between the rates.
dV/dt = A*dh/dt</span>
<span> in the meantime the area is not altering </span>
<span>
dV/dt = π*(1 ft)^2*(-0.5 ft/min) </span>
<span>
dV/dt = -π/2 ft^3/min ≈ -1.571 ft^3/min
Water is draining from the tank at the rate of π/2
ft^3/min.</span>
