Answer:
Explanation:
given data:
height of tank = 60cm
diameter of tank =40cm
accelration = 4 m/s2
suppose x- axis - direction of motion
z -axis - vertical direction
= water surface angle with horizontal surface
accelration in x direction
accelration in z direction
slope in xz plane is
the maximum height of water surface at mid of inclination is
the maximu height of wwater to avoid spilling is
= 60 - 8.2
the height requird if no spill water is
Answer:
... spilling water or getting anything cascading onto the floor
Answer:
HUMAN DEVELOPMENT
MOTOR BEHAVIOR
EXERCISE SCIENCE
MEASUREMENT AND EVALUATION
HISTORY AND PHILOSOPHY
UNIQUE ATTRIBUTES OF LEARNERS
CURRICULUM THEORY AND DEVELOPMENT
Explanation:
The correct question;
An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?
Answer:
h'_2 = 40 W/K.m²
Explanation:
We are given;
L1 = 1m
L2 = 5m
T_s = 400 K
T_(∞) = 300 K
V = 100 m/s
q = 20,000 W/m²
Both objects have the same shape and density and thus their reynolds number will be the same.
So,
Re_L1 = Re_L2
Thus, V1•L1/v1 = V2•L2/v2
Hence,
(h'_1•L1)/k1 = (h'_2•L2)/k2
Where h'_1 and h'_2 are convection coefficients
Since k1 = k2, thus, we now have;
h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)
Thus,
h'_2 = [20,000/(400 - 300)]•(1/5)
h'_2 = 40 W/K.m²
Answer:
The percentage of the remaining alloy would become solid is 20%
Explanation:
Melting point of Cu = 1085°C
Melting point of Ni = 1455°C
At 1200°C, there is a 30% liquid and 70% solid, the weight percentage of Ni in alloy is the same that percentage of solid, then, that weight percentage is 70%.
The Ni-Cu alloy with 60% Ni and 40% Cu, and if we have the temperature of alloy > temperature of Ni > temperature of Cu, we have the follow:
60% Ni (liquid) and 40% Cu (liquid) at temperature of alloy
At solid phase with a temperature of alloy and 50% solid Cu and 50% liquid Ni, we have the follow:
40% Cu + 10% Ni in liquid phase and 50% of Ni is in solid phase.
The percentage of remaining alloy in solid is equal to
Solid = (10/50) * 100 = 20%
