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3 years ago

The costs of mining and transporting coal are roughly independent of the heating value of the coal. Consider:

Engineering

1 answer:

Paul [167]3 years ago

5 0

Answer:

Explanation:

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A 75 mm diameter, 15 mm wide A80-M12V grinding wheel is used for traverse surface grinding of a 75 mm by 75 mm 1040 steel workpi

Dmitrij [34]

Answer:

Detailed solution is given and calculations are done

3 0

3 years ago

A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard sea-level air, while theprototype

ella [17]

Answer:

Explanation:

Given

scale i.e. $L_r=1:15$

Using Reynolds number similarity

$(Re)_m=(Re)_p$

$(\frac{Vl}{\nu })_m=(\frac{Vl}{\nu })_p$

Properties of air

$\nu _{air}=1.57\times 10^{-4} ft/s$

Properties of sea water

$\nu _{sea}=1.26\times 10^{-5} ft/s$

$\left ( \frac{V_ml_M}{\nu _m}\right )=\left ( \frac{V_pl_p}{\nu _p}\right )$

$V_p=V_m\left ( \frac{l_m}{l_p}\right )\left ( \frac{\nu _p}{\nu _m}\right )$

$V_p=180\times \frac{1}{15}\times \frac{1.26\times 10^{-5}}{1.57\times 10^{-4}}$

$V_p=0.963\ ft/s$

8 0

3 years ago

What tools do you need to change a flat tire

Lyrx [107]

Answer:

Spare tire (if you’re dealing with a flat and not just rotating tires or accessing the brakes)
Carjack
Lug wrench
Wheel wedges
Work gloves
Tire repair kit (if you’re attempting to fix your flat tire)
Flare/reflective triangles (if you’re changing the tire on the side of the road or in a parking lot)

6 0

4 years ago

Read 2 more answers

If an object is near the surface of the earth the variation of its weight with distance from the center of the earth can ofteb b

zalisa [80]

Answer:

h=32.1 km

Explanation:

<em>solution:</em>

using newton law of gravitational attraction and newton second law:

$W=\frac{Gmm_{E} }{r^{2} } \\a=\frac{Gm_{E}}{r^{2}} \\W=ma\\$

$m_{E}= mass of earth$

r= distance between two masses

at sea level

a=g

$r=R_{E}$

$a=\frac{Gm_{E}}{r^{2}}$ .............................(1)

$Gm_{E} =gR_{E}^2$ .........................(2)

by substituting (2) and (1) $a=g\frac{R_{E}^2 }{r^{2} }$ acceleration due to gravity at a distance r from the centre of the earth in terms of g (sea level)

so the weight of the object at a distance r from the centre of the earth (W=ma)

W=mg(Re^2/r^2)..........(3)

h the height above the surface of the earth: r=Re+h

putting the value of r in eq (3)

W=mg(Re/Re+h)^2

W=0.99 mg

solving for height h:

h=Re(1/√0.99)-(1))

h=32.1 km

4 0

3 years ago

CAN YALL GO FOLLOW MY INSTAGRAM PLSS

lions [1.4K]

Answer:

what is it

Explanation:

7 0

3 years ago

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