Answer:
<u><em>both, one</em></u>
Explanation:
<em>Alternating current flows in both directions and direct current flows in one direction.</em>
<em></em>
<em>Hope it helps.</em>
<em>;)</em>
<em><3</em>
Answer:
a. 430.944 pascal
b. 0.0625psi
c. 1.73008inH20
Explanation:
The pressure rise Ap associated with wind hitting a win- dow of a building can be estimated using the formula Ap-p(12/2), where p is density of air and V is the speed of the wind. Apply the grid method to calculate pressure rise for P-1.2 kg/m and V-60 mph. a. Express your answer in pascals. b. Express your answer in pounds-force per square inch (psi). c. Express your answer in inches of water column
checking the dimensional consistency
Dp=

convert 1 mile to meter
1mile=1609m
1h=3600s
60mile/h=26.8m/s
slotting intpo the relation

430.944kg/(ms^2)
which is the same as 430.944N/m^2
expressing in pascal.
We know that
1 pascal=1 N/m^2
430.944 pascal
2. 1 pascal=0.000145psi
answer=0.0625psi
3.1 pascal=0.00401inH20
answer=430.944*0.0040146
1.73008inH20
Answer:
Space mean speed = 44 mi/h
Explanation:
Using Greenshield's linear model
q = Uf ( D -
/Dj )
qcap = capacity flow that gives Dcap
Dcap = Dj/2
qcap = Uf. Dj/4
Where
U = space mean speed
Uf = free flow speed
D = density
Dj = jam density
now,
Dj = 4 × 3300/55
= 240v/h
q = Dj ( U -
/Uf)
2100 = 240 ( U -
/55)
Solve for U
U = 44m/h
Answer:
Statement 1: All balls hit the ground at the same time
Explanation:
When there is no resistance of air, the acceleration due to gravity experienced by all the bodies are same. So for falling bodies, neglecting the air resistance, the falling object will be weightless and therefore all the objects will hit the ground at the same time when there is nor air resistance and the objects are considered to be falling in vacuum.