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Butoxors [25]
3 years ago
9

In the pressure filled driving environment, handicapping yourself by drinking and driving is not a good choice.

Engineering
1 answer:
vodka [1.7K]3 years ago
6 0

Answer:

True

Explanation:

Please mark me brainlist

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You have a motor that runs at 1.5 amps from a 12 volt power source, how many watts of power is it using?
Leviafan [203]

Answer:

18 Watts

Explanation:

For this problem, we simply need to understand the relationship of power to voltage and current.  This relationship is derived from Ohm's law:

Power = Voltage * Current

Given this equation, we can say the following to find the power consumption of the motor:

Power = 12volts * 1.5amps

Power = 18 Watts

Hence, the motor is consuming 18 Watts of power.

Cheers.

3 0
3 years ago
Identify three questions a patient might ask of the nuclear medicine technologist performing a nuclear medicine exam.
bogdanovich [222]

Answer:

How long is my nuclear medicine exam?

How long will the radioactivity stay in my system?

What are the risks?

Explanation:

3 0
2 years ago
1 Define Engineering <br><br> 2 Engineering use a (n) _______ process to solve problems
OverLord2011 [107]
Engineering use a (n) Engineering design process to solve problems.

4 0
3 years ago
Consider the brass alloy for which the stress-strain behavior is shown in the Animated Figure 7.12. A cylindrical specimen of th
sleet_krkn [62]

Answer:

(a) 0.1509 mm

(b) 0.00525 mm

Explanation:

Stress, \sigma is given by

\sigma=\frac {F}{A} where F is force and A is area and area is given by \frac {\pi d^{2}}{4} hence

\sigma=\frac {4F}{\pi d^{2}} where d is the diameter. Substituting 9970 N for F and 10mm=0.01 m for d hence

\sigma=\frac {4*9970 N}{\pi 0.01m^{2}}=126941982.6 N/m^{2}

\sigma \approx 127 Mpa

From the attached stress-strain diagram, the stress of 127 Mpa corresponds to strain of 0.0015 and since strain is given by

\epsilon=\frac {\triangle l}{l} where\epsilon is the strain, \triangle l is elongation and l is original length and making elongation the subject

\triangle l= \epsilon \times l and substituting strain with 0.0015 and length l with 100.6 mm then

\triangle l=0.0015\times 100.6=0.1509 mm

(b)

Lateral strain is given by

\epsilon_{lat}=\frac {\triangle d}{d} and substituting -v\epsilon for \epsilon_{lat} where v is poisson ratio then

-v\epsilon=\frac {\triangle d}{d} and making \triangle d the subject then

\triangle d=-vd\epsilon and substituting 0.35 for v, 0.0015 for strain and 10 mm for d

\triangle d=-(0.35)*10*0.0015=-0.00525 mm and the negative sign indicates decrease in diameter

8 0
3 years ago
Consider a 0.15-mm-diameter air bubble in a liquid. Determine the pressure difference between the inside and outside of the air
777dan777 [17]

Answer:

a)\Delta P= 2666.66 Pa

b)\Delta P= 3200 Pa

Explanation:

Given that

Diameter ,d= 0.15 mm

We know that pressure difference is given as

\Delta P=\dfrac{4\sigma }{d}

Now by putting the values

When surface tension 0.1 N/m  :

The surface tension ,\sigma=0.1\ N/m

\Delta P=\dfrac{4\times 0.1 }{0.15}\times 10^3\ Pa

\Delta P= 2666.66 Pa

When surface tension 0.12 N/m  :

The surface tension ,\sigma=0.12\ N/m

\Delta P=\dfrac{4\times 0.12 }{0.15}\times 10^3\ Pa

\Delta P= 3200 Pa

a)\Delta P= 2666.66 Pa

b)\Delta P= 3200 Pa

4 0
4 years ago
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