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Anettt [7]
3 years ago
14

How high a building could fire hoses effectively spray from the ground? Fire hose pressures are around 1 MPa. (It is also said t

hat larger fire engine mounted pumps can deliver from 90-300 gallons per minute, and hoses can be as wide as 30 cm diameter What alternative estimate would that info give you?)
Engineering
1 answer:
Mrac [35]3 years ago
7 0

Answer:

z_{2} = 91.640\,m

Explanation:

The phenomenon can be modelled after the Bernoulli's Principle, in which the sum of heads related to pressure and kinetic energy on ground level is equal to the head related to gravity.

\frac{P_{1}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}= z_{2}+\frac{P_{2}}{\rho\cdot g}

The velocity of water delivered by the fire hose is:

v_{1} = \frac{(300\,\frac{gal}{min} )\cdot(\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot (0.3\,m)^{2}}

v_{1} = 0.267\,\frac{m}{s}

The maximum height is cleared in the Bernoulli's equation:

z_{2}= \frac{P_{1}-P_{2}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}

z_{2}= \frac{1\times 10^{6}\,Pa-101.325\times 10^{3}\,Pa}{(1000\,\frac{kg}{m^{3}} )\cdot(9.807\,\frac{m}{s^{2}} )} + \frac{(0.267\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}

z_{2} = 91.640\,m

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An automated transfer line is to be designed. Based on previous experience, the average downtime per occurrence = 5.0 min, and t
IRINA_888 [86]

Answer:

a) 28 stations

b) Rp = 21.43

E = 0.5

Explanation:

Given:

Average downtime per occurrence = 5.0 min

Probability that leads to downtime, d= 0.01

Total work time, Tc = 39.2 min

a) For the optimum number of stations on the line that will maximize production rate.

Maximizing Rp =minimizing Tp

Tp = Tc + Ftd

=  \frac{39.2}{n} + (n * 0.01 * 5.0)

= \frac{39.2}{n} + (n * 0.05)

At minimum pt. = 0, we have:

dTp/dn = 0

= \frac{-39.2}{n^2} + 0.05 = 0

Solving for n²:

n^2 = \frac{39.2}{0.05} = 784

n = \sqrt{784} = 28

The optimum number of stations on the line that will maximize production rate is 28 stations.

b) Tp = \frac{39.2}{28} + (28 * 0.01 * 5)

Tp = 1.4 +1.4 = 2.8

The production rate, Rp =

\frac{60min}{2.8} = 21.43

The proportion uptime,

E = \frac{1.4}{2.8} = 0.5

3 0
3 years ago
Without motorcycle riders are at risk of severe injury in a crash ?
nekit [7.7K]

Answer:

The correct option is;

A. proper protection

Explanation:

Motorcycle riders ride the motorcycle while at some level of speed while  having the entire body exposed to be a major part of any collision.

Injuries sustained from motorcycle accidents are several times more severe than injuries sustained by occupants of a car that is fully protected by the metallic panel in the same and even more serious accident scenarios

Hence, motorcycle riders require adequate protection by putting on available motorcyclist safety gear

Therefore, to reduce the risk of severe injury n a crash, motorcycle riders require proper protection.

5 0
3 years ago
Read 2 more answers
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Scorpion4ik [409]
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3 0
3 years ago
Read 2 more answers
Determine the period of each of the following discrete-time signals (if a signal is not periodic, denote its period by infinity)
sergiy2304 [10]

Answer:

a) it is periodic

N = (20/3)k = 20 { for K =3}

b) it is Non-Periodic.

N = ∞

c) x(n) is periodic

N = LCM ( 5, 20 )

Explanation:

We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.

then the period of the signal is given as

N = ( 2π/w₀)K

k is least integer for which N is also integer

Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2

now

a) cos(2π(0.15)n)

w₀ = 2π(0.15)

Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3

so, it is periodic

N = (20/3)k = 20 { for K =3}

b) cos(2n);

w₀ = 2

Now, 2π/w₀ = 2π/2) = π

so, it is Non-Periodic.

N = ∞

c)  cos(π0.3n) + cos(π0.4n)

x(n) = x1(n) + x2(n)

x1(n) = cos(π0.3n)

x2(n) =  cos(π0.4n)

so

w₀ = π0.3

2π/w₀  = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3

∴ N1 = 20

AND

w₀ = π0.4

2π/w₀  = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5

∴ N² = 5

so, x(n) is periodic

N = LCM ( 5, 20 )

6 0
3 years ago
Express (118)10 and (-49)10 in 8-bit binary one’s complement form and then add the numbers. What would be the representation (-0
stealth61 [152]

Answer:

118_{10}= 0110111 2) -49_{10}=110001_{2} 3) 0_{10}=0:16 \Rightarrow 0_{10}=0_{16}

Explanation:

1) Expressing the Division as the summation of the quotient and the remainder

for

118, knowing it is originally a decimal form:

118:2=59  +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

118_{10}= 0110111

2) -49_{10}

Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.

|-49|=49

49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1

100011

-49_{10}=110001_{2}

3) (-0)_{10}

The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put:

(-0)_{10}=0:16=0 \Rightarrow (0)_{10}=0_{16} \:or\\(0)_{16}=0000000000000000

5 0
3 years ago
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