Less, if it’s too big: hard to control and maneuverability for shooting wouldn’t be that good. a smaller wheelchair allows for faster movement and control, along with easier shooting and upper body movement
Answer:
a) 0.01
b) 150 cm^3/s
c) 300 cm^3/s
d) 25 cm^3/s
Explanation:
a) We know that :
Q=ΔP/R
R=8ηl/π*r^4
Givens:
r^2 = 0.1 r_1
Plugging known information to get :
Q=ΔP/R
=ΔP*π*r^4/8*η*l
Q_2/r_2^4 =Q_1/r_1^4
Q_2=Q_1/r_1^4*r_2^4
=Q_1/r_1^4*r*0.0001*r_1^4
Q_2 = 0.01
b) From the rate flow of the fluid we know that :
Q=ΔP/R (1)
F=η*Av/l (2)
R=8*ηl/π*r^4 (3)
<em>Where: </em>
ΔP is the change in the pressure .
r is the raduis of the tube .
l is the length of the tube .
η is the coefficient of the vescosity of the fluid .
R is the resistance of the fluid .
Givens: Q1 = 100 cm^3/s , ΔP= 1.5
Plugging known information into EQ.1 :
Q=ΔP/R
Q_2/ΔP2=Q_1/ΔP
Q_2=150 cm^3/s
c) we know that :
F = η*Av/l
can be written as :
ΔP = F/A = η*v/l
Givens: η_2 = 3η_1
Q=ΔP/R
Q=η*v/l*R
Q_2/η_2=Q_1/η_1
Q_2=300 cm^3/s
d) We know that :
Q=ΔP/R
R=8*ηl/π*r^4
Givens: l_2 = 4*l_1
Plugging known information to get :
Q=ΔP/R
Q=ΔP*π*r^4/8*ηl
Q_2/l_2=Q_1/l_1
Q_2 = 25 cm^3/s
Answer:
center left-turn lane
Explanation:
A <em>center left turn lane</em> will be marked as described. The arrows, if present, generally indicate that left turns are permitted from the lane with these markings.
__
If the double yellow lines are solid, they are considered to be a "barrier" and are not to be crossed.
Answer:
The Full details of the answer is attached.
Answer:
so heat loss = 4312 W
cost of heat loss daily is $8.28 per day
Explanation:
given data
slab length L = 11 m
slab wide W = 8 m
thickness t = 0.20 m
temperature top T1 = 17°C
temperature bottom T2 = 10°C
thermal conductivity k = 1.4 W/m-K
efficiency ηf = 0.90
priced Cg = $0.02 / MJ
to find out
rate of heat loss and daily cost of the heat loss
solution
we calculate here heat loss by heat transfer
so apply here formula that is
q = (thermal conductivity × area × temperature difference) / thickness
put here all these value we get heat loss
q =
q =
q = 4312 W
so heat loss = 4312 W
and
cost of the heat loss is express as
cost of heat loss =
put here all these value
cost of heat loss is = × 24 hr/day × 3600 s/hr
cost of heat loss = 8.279
so cost of heat loss daily is $8.28 per day