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3 years ago

What is the minimum recommended safe distance from an X-ray source?

Engineering

2 answers:

lara31 [8.8K]3 years ago

3 0

Hi! I believe the answer is 2 meters(:

otez555 [7]3 years ago

3 0

The answer is 2mater

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What are some advantages of making electronic components like transistors increasingly smaller? (40 POINTS) Pls help

Karo-lina-s [1.5K]

it makes it so that its much easier to have these objects around,like how when you wear the translator its much more simple,and instant.

8 0

3 years ago

A heavy-duty electrical resistor is 2cm in diameter by 16cm long. 5 amps of current through it heats the resistor to 100°C, and

Olenka [21]

Answer:

The convective coefficient is 37.3 W/m²K.

Explanation:

Use Newton’s law of cooling to determine the heat transfer coefficient. Assume there is no heat transfer from the ends of electric resistor. Heat is transferred from the resistor curved surface.

Step1

Given:

Diameter of the resistor is 2 cm.

Length of the resistor is 16 cm.

Current is 5 amp.

Voltage is 6 volts.

Resistor temperature is 100°C.

Room air temperature is 20°C.

Step2

Electric power from the resistor is transferred to heat and this heat is transferred to the environment by means of convection.

Power of resistor is calculated as follows:

P=VI

$P=6\times5$

P= 30 watts.

Step3

Newton’s law of cooling is expressed as follows:

$Q=h\times \pi DL(T_{r}-T_{\infty})$

Here, h is the convection heat coefficient and $\pi DL$ is the exposed surface area of the resistor.

Substitute the values as follows:

$30=h\times \pi (2cm)(\frac{1m}{100cm})(16cm)(\frac{1m}{100cm})(100-20)$

$h=\frac{30}{0.8042}$

h = 37.3 W/m²K.

Thus, the convective coefficient is 37.3 W/m²K.

6 0

3 years ago

A pump is used to circulate hot water in a home heating system. Water enters the well-insulated pump operating at steady state a

Roman55 [17]

Answer:

Temperature change is 0.0345 degree rankine

The change in temperature is therefore small

Explanation:

Please see attached for the workings.

8 0

4 years ago

Tool life testing on a lathe under dry cutting conditions gauge 'n' and 'C' of Taylor tool life equation as 0.12 and 130 m/min.

Tom [10]

Answer:

So % increment in tool life is equal to 4640 %.

Explanation:

Initially n=0.12 ,V=130 m/min

Finally C increased by 10% , V=90 m/min

Let's take the tool life initial condition is $T_1$ and when C is increased it become $T_2$ .

As we know that tool life equation for tool

$VT^n=C$

At initial condition $130\times (T_1)^{0.12}=C$ ------(1)

At final condition $90\times (T_2)^{0.12}=1.1C$ -----(2)

From above equation

$\dfrac{130\times (T_1)^{0.12}}{90\times (T_2)^{0.12}}=\dfrac{1}{1.1}$

$T_2=47.4T_1$

So increment in tool life = $\dfrac{T_2-T_1}{T_1}$

= $\dfrac{47.4T_1-T_1}{T_1}$

So % increment in tool life is equal to 4640 %.

7 0

3 years ago

The electricity generated by wind turbines annually in kilowatt-hours per year is given in a file. The amount of electricity is

Anika [276]

Answer:

Steps:

1. Create a text file that contains blade diameter (in feet), wind velocity (in mph) and the approximate electricity generated for the year

2. load the data file for example, in matlab, use ('fileame.txt') to load the file

3. create variables from each column of your data

for example, in matlab,

x=t{1}

y=t{2}

4. plot the wind velocity and electricity generated.

plot(x, y)

5. Label the individual axis and name the graph title.

title('Graph of wind velocity vs approximate electricity generated for the year')

xlabel('wind velocity')

ylabel('approximate electricity generated for the year')

5 0

3 years ago

Read 2 more answers