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3 years ago

How would you design a wheelchair for wheelchair-using basketball players? Would you make it more or less massive?

1 answer:

6 0

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

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A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i

dlinn [17]

Answer:

$h_{max} = 51.8 cm$

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

$\theta$ = water surface angle with horizontal surface

$a_x =$ accelration in x direction

$a_z =$ accelration in z direction

slope in xz plane is

$tan\theta = \frac{a_x}{g +a_z}$

$tan\theta = \frac{4}{9.81+0}$

$tan\theta =0.4077$

the maximum height of water surface at mid of inclination is

$\Delta h = \frac{d}{2} tan\theta$

$=\frac{0.4}{2}0.4077$

$\Delta h 0.082 cm$

the maximu height of wwater to avoid spilling is

$h_{max} = h_{tank} -\Delta h$

= 60 - 8.2

$h_{max} = 51.8 cm$

the height requird if no spill water is $h_{max} = 51.8 cm$

3 0

4 years ago

2. A counter flow tube-shell heat exchanger is used to heat a cold water stream from 18 to 78oC at a flow rate of 1 kg/s. Heatin

Anastaziya [24]

Answer:

a) $L = 220\,m$ , b) $U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

Explanation:

a) The counterflow heat exchanger is presented in the attachment. Given that cold water is an uncompressible fluid, specific heat does not vary significantly with changes on temperature. Let assume that cold water has the following specific heat:

$c_{p,c} = 4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C}$

The effectiveness of the counterflow heat exchanger as a function of the capacity ratio and NTU is:

$\epsilon = \frac{1-e^{-NTU\cdot(1-c)}}{1-c\cdot e^{-NTU\cdot (1-c)}}$

The capacity ratio is:

$c = \frac{C_{min}}{C_{max}}$

$c = \frac{(1\,\frac{kg}{s} )\cdot(4.186\,\frac{kW}{kg^{\textdegree}C} )}{(1.8\,\frac{kg}{s} )\cdot(4.30\,\frac{kW}{kg^{\textdegree}C} )}$

$c = 0.541$

Heat exchangers with NTU greater than 3 have enormous heat transfer surfaces and are not justified economically. Let consider that $NTU = 2.5$ . The efectiveness of the heat exchanger is:

$\epsilon = \frac{1-e^{-(2.5)\cdot(1-0.541)}}{1-(2.5)\cdot e^{-(2.5)\cdot (1-0.541)}}$

$\epsilon \approx 0.824$

The real heat transfer rate is:

$\dot Q = \epsilon \cdot \dot Q_{max}$

$\dot Q = \epsilon \cdot C_{min}\cdot (T_{h,in}-T_{c,in})$

$\dot Q = (0.824)\cdot (4.186\,\frac{kW}{^{\textdegree}C} )\cdot (160^{\textdegree}C-18^{\textdegree}C)$

$\dot Q = 489.795\,kW$

The exit temperature of the hot fluid is:

$\dot Q = \dot m_{h}\cdot c_{p,h}\cdot (T_{h,in}-T_{h,out})$

$T_{h,out} = T_{h,in} - \frac{\dot Q}{\dot m_{h}\cdot c_{p,h}}$

$T_{h,out} = 160^{\textdegree}C + \frac{489.795\,kW}{(7.74\,\frac{kW}{^{\textdegree}C} )}$

$T_{h,out} = 96.719^{\textdegree}C$

The log mean temperature difference is determined herein:

$\Delta T_{lm} = \frac{(T_{h,in}-T_{c, out})-(T_{h,out}-T_{c,in})}{\ln\frac{T_{h,in}-T_{c, out}}{T_{h,out}-T_{c,in}} }$

$\Delta T_{lm} = \frac{(160^{\textdegree}C-78^{\textdegree}C)-(96.719^{\textdegree}C-18^{\textdegree}C)}{\ln\frac{160^{\textdegree}C-78^{\textdegree}C}{96.719^{\textdegree}C-18^{\textdegree}C} }$

$\Delta T_{lm} \approx 80.348^{\textdegree}C$

The heat transfer surface area is:

$A_{i} = \frac{\dot Q}{U_{i}\cdot \Delta T_{lm}}$

$A_{i} = \frac{489.795\,kW}{(0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C} )\cdot(80.348^{\textdegree}C) }$

$A_{i} = 9.676\,m^{2}$

Length of a single pass counter flow heat exchanger is:

$L =\frac{A_{i}}{\pi\cdot D_{i}}$

$L = \frac{9.676\,m^{2}}{\pi\cdot (0.014\,m)}$

$L = 220\,m$

b) Given that tube wall is very thin, inner and outer heat transfer areas are similar and, consequently, the cold side heat transfer coefficient is approximately equal to the hot side heat transfer coefficient.

$U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

5 0

3 years ago

The first step in treating shock is to

san4es73 [151]

Answer:

Lay the person down and elevate thier legs slightly.

Explanation:

5 0

3 years ago

How can I solve 23.5 million Nona meters to millimeters using no calculator because I have to show my work

lozanna [386]

Answer:

its so simple. u must mind some formulas.

Explanation:

mili->10^(-3)

micro->10^(-6)

nano->10^(-9)

so write the exact number and move "." to left or right depend question.

in this one:

23.5 is 23500000.0 nano with a default dot at the end

for turning to mili u must move the dot 6 steps to left so it will be: 23.5 milimeter.

6 0

3 years ago

Astronauts who landed on the moon during the Apollo 15, 16, and 17 missions brought back a large collection of rocks to the eart

stiks02 [169]

Answer: a) W(earth) = 935.62 lbs

b) Mass of rocks in slugs = 29.06 slugs

Explanation:

a) From Newton's law, W = mg. Whether on the moon or on earth. Although, the mass of the rocks everywhere is the same, that is, mass of rocks on the moon = mass of rocks on earth.

W(moon) = mg(moon)

W(moon) = 154 lbs

g(moon) = 5.30 ft/s2

m = W(moon)/g(moon) = 154/5.3 = 29.06 lb.s2/ft

W(earth) = m g(earth)

g(earth) = 32.2 ft/s2

W(earth) = 29.06 × 32.2 = 935.62 lbs.

b) A slug = 1 lb.s2/ft, therefore the mass of the rocks in slugs is 29.06 slugs.

QED!

8 0

3 years ago