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3 years ago

The author uses the example of the flameless candle to illustrate that

Engineering

2 answers:

Vesnalui [34]3 years ago

6 0

Answer:

Engineering makes products safer

Explanation:

krek1111 [17]3 years ago

4 0

Answer:

I don't necessarily know the context of the text you are talking about, but a flameless candle could represent how there no more light in the author's life, which could represent a lack of happiness or a loss. Hope this helps!

Explanation:

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To solve the problem, make assumptions for missing data and justify. Given:

finlep [7]

Answer:

5,4,1, this is a explication

6 0

2 years ago

Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o

jekas [21]

Complete Question

Complete Question is attached below.

Answer:

$V'=5m/s$

Explanation:

From the question we are told that:

Diameter $d=0.10m$

Power $P=4.0kW$

Head loss $\mu=10m$

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu$

$\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10$

$H_m=(\frac{200*10^3}{1000*9.8}-10)$

$H_m=10.39m$

Generally the equation for Power is mathematically given by

$P=\rho gQH_m$

Therefore

$Q=\frac{P}{\rho g H_m}$

$Q=\frac{4*10^4}{1000*9.81*10.9}$

$Q=0.03935m^3/sec$

Since

$Q=AV'$

Where

$A=\pi r^2\\A=3.142 (0.05)^2$

$A=7.85*10^{-3}$

Therefore

$V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}$

$V'=5m/s$

5 0

3 years ago

Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De

jok3333 [9.3K]

Answer:

$Q=4.98\times 10^{-3}\ m^3/s$ .

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

copper tube is 3/4 standard type K drawn tube.

From standard chart ,the dimension of 3/4 standard type K copper tube given as

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

$\mu =0.00164\ Pa.s$

We know that

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

Where Q is volume flow rate

L is length of tube

$d_i$ is inner diameter of tube

ΔP is pressure drop

μ is dynamic viscosity

Now by putting the values

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

$130\times 1000=\dfrac{128\times 0.00164\times 50Q}{\pi \times 0.01892^4}$

$Q=4.98\times 10^{-3}\ m^3/s$

So flow rate is $Q=4.98\times 10^{-3}\ m^3/s$ .

6 0

3 years ago

Steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a

Veronika [31]

Answer:

a. 6 seconds

b. 180 feet

Explanation:

Images attached to show working.

a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero is when it stops.

b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.

7 0

3 years ago

How deep is a 6ft hole?

Pavlova-9 [17]

Answer:

I know this sounds quite deep but it is as deep as a grave

Explanation:

It's reality

3 0

2 years ago

Read 2 more answers

The author uses the example of the flameless candle to illustrate that​

The author uses the example of the flameless candle to illustrate that