Answer:
A) At point 1, local acceleration = 0.5 m/s²
At point 2, local acceleration = 1.0 m/s²
B) Average Eulerian convective acceleration over the two points in the cross section shown = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
Explanation:
Local acceleration at those points is the instantaneous acceleration at those points and it is given as
a = dv/dt
At point 1, v₁ = 0.5 t
a₁ =dv₁/dt = 0.5 m/s²
At point 2, v₂ = 1.0 t
a₂ = dv₂/dt = 1.0 m/s²
b) Average Eulerian convective acceleration over the two points in the cross section shown = (change of velocity between the two points)/time
Change of velocity between the two points = v₂ - v₁ = 1.0t - 0.5t = 0.5 t
Time = t
Average acceleration = 0.5t/t = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
Answer:
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Explanation:
I really need this points
Answer:
100°c = 373.15 K
100°C=212°F
Explanation:
To convert Celsius to Kelvin, we need the following equation.
°C + 273.15 = K
100°C + 273.15 = K
373.15 = K
Therefore, 100°c = 373.15 K
F = 9/5C + 32
=9/5(100)+32
= (180) + 32
= 212°
Therefore,
100°C=212°F
13.0m/s
1.2m/s
Explanation:
Given parameters:
Initial speed of the body = 7.1m/s
time taken = 2.23s
Acceleration = 2.64m/s²
Unknown:
Final speed = ?
Solution:
Acceleration is the rate of change of velocity with time.
a = 
a = acceleration
V = final speed
U = initial speed
T = time taken
Input the variables and solve for V;
2.64 =
V - 7.1 = 5.9 expression 1
V = 5.9 + 7.1 = 13.0m/s
B
Using the same parameters, the speed after a uniform deceleration of -2.64m/s², the negative sign implies deceleration;
from expression 1;
V - 7.1 = -5.9
V = -5.9 + 7.1 = 1.2m/s
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Answer:
this pdf should help you out
Explanation: