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2 years ago

Identify three questions a patient might ask of the nuclear medicine technologist performing a nuclear medicine exam.

Engineering

1 answer:

bogdanovich [222]2 years ago

3 0

Answer:

How long is my nuclear medicine exam?

How long will the radioactivity stay in my system?

What are the risks?

Explanation:

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a coil consists of 200 turns of copper wire and has a cross-sectional area of 0.8mm square . The mean length per turn is 80 cm a

Amanda [17]

Answer:

The picture below with the answer. Hope it helps, have a great day/night and stay safe! Length of the coil,

8 0

3 years ago

Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of

earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant$

Since the losses are neglected thus applying this theorm between upper and lower porion we have

$\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}$

Now by continuity equation we have

$A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s$

Applying the values in the Bernoulli's equation we get

$\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars$

6 0

3 years ago

The speed of sound in a fluid can be calculated using the following equation:

wlad13 [49]

Answer:

Jesus is always the answer

4 0

2 years ago

An inventor claims to have invented a heat engine that operates between the temperatures of 627°C and 27°C with a thermal effici

Oksana_A [137]

Answer Explanation:

the efficiency of the the engine is given by=1- $\frac{T_2}{T_1}$

where T₂= lower temperature

T₁= Higher temperature

we have given efficiency =70%

lower temperature T₂=27°C=273+27=300K

higher temperature T₁=627°C=273+627=900K

efficiency=1- $\frac{T_2}{T_1}$

=1- $\frac{300}{900}$

=1-0.3333

=0.6666

=66%

66% is less than 70% so so inventor claim is wrong

3 0

3 years ago

Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat

zlopas [31]

Answer: $10.631\times 10^3\ N/m^2$

Explanation:

Given

Discharge is $Q=12.5\ L$

Diameter of pipe $d=150\ mm$

Distance between two ends of pipe $L=800\ m$

friction factor $f=0.008$

Average velocity is given by

$\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s$

Pressure difference is given by

$\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa$

8 0

2 years ago