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LiRa [457]
3 years ago
15

1. Calculate the battery life in years when a pacemaker has the following characteristics: Battery Ampere-hours = 1.5 Pulse volt

age = 2V Pulse width = 1.5 msec Pulse time period = 1 sec Electrode heart resistance = 150Ω Current drain on the battery = 1.25 µA
Engineering
1 answer:
Wittaler [7]3 years ago
4 0

Answer:

battery life in year = 9 years and 48 days

Explanation:

given data

Battery Ampere-hours = 1.5

Pulse voltage = 2 V

Pulse width = 1.5 m sec

Pulse time period = 1 sec

Electrode heart resistance = 150 Ω

Current drain on the battery = 1.25 µA

to find out

battery life in years

solution

we get first here duty cycle that is express as

duty cycle = \frac{width}{period}      ...............1

duty cycle = 1.5 × 10^{-3}

and applied voltage will be

applied voltage = duty energy × voltage    ...........2

applied voltage = 1.5 × 10^{-3} × 2

applied voltage = 3 mV

so current will be

current = \frac{applied\ voltage}{resistance}   ................3

current = \frac{3}{150}

current = 20 µA

so net current will be

net current = 20 - 1.25

net current = 18.75 µA

so battery life will be

battery life = \frac{1.5}{18.75*10^{-6}}

battery life = 80000 hours

battery life in year = \frac{80000}{8760}

battery life in year = 9.13 years

battery life in year = 9 years and 48 days

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oksano4ka [1.4K]

Answer:

a)A constant volume process is called isochoric process.

b)Yes

c)Work =0

Explanation:

Isochoric process:

 A constant volume process is called isochoric process.

In constant volume process work done on the system or work done by the system will remain zero .Because we know that work done give as

work = PΔV

Where P is pressure and ΔV is the change in volume.

For constant volume process ΔV = 0⇒ Work =0

Yes heat transfer can be take place in isochoric process.Because we know that temperature difference leads to transfer of heat.

Given that

Initial P=10 MPa

Final pressure =15 MPa

Volume = 100 L

Here volume of gas is constant so the work work done will be zero.

4 0
4 years ago
While walking across campus one windy day, an engineering student speculates about using an umbrella as a "sail" to propel a bic
makvit [3.9K]

Answer:

Given data:\\While walking across campus one windy day\\Frontal area, \(A=0.3 m ^{2}\)\\Wind speed \(V=24 Km / hr\)\\The drag coefficient \(C_{D, b}=1.2\)\\The combined mass \(m=75 kg\)\\Umbrella diameter, \(D=1.22 m\)\\Velocity of wind \(V=24 \frac{ km }{ hr }\)\\The rolling resistance \(C_{R}=0.75 \%\)

Solution:

Note: Refer the diagram

Basic equation:\\'s law of motion: \(\sum F_{x}=m a_{x}\)\\Lift coefficient, \(C_{L}=\frac{F_{L}}{\frac{1}{2} \rho V^{2} A_{p}}\)\\Drag coefficient, \(C_{D}=\frac{F_{D}}{\frac{1}{2} \rho V^{2} A_{p}}\)

From force balance equation:\\\(\sum F_{x}=F_{D}-F_{R}=0\)\\But \(F_{D}=\left(C_{D, \alpha} A_{u}+C_{D, B} A_{b}\right) \frac{1}{2} \rho\left(V_{\nu}-V_{b}\right)^{2}\)\(F_{R}=C_{R} m g\)\\Area of the Umbrella \(A_{u}=\frac{\pi D_{u}^{2}}{4}\)\(A_{x}=\frac{\pi \times 1.22^{2}}{4} m ^{2}\)\(A_{v}=1.17 m ^{2}\)

Drag coefficient data for selected objects table at

Hemisphere (open end facing flow), C_{D, x}=1.42

Substituting all parameters,

\begin{aligned}&F_{R}=0.0075 \times 75 \times 9.81\\&F_{R}=5.52 N\end{aligned}

Then,

\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}

And the equation becomes,

\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}

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7 0
4 years ago
Consider a piston-cylinder device with a piston surface area of 0.1 m^2 initially filled with 0.05 m^3 of saturated water vapor
miv72 [106K]

The friction force f = 10000 N

The heat transfer Q = 1.7936 KJ

<u>Explanation:</u>

Given data:

Surface area of Piston = 1 m^{2}

Volume of saturated water vapor = 100 K Pa

Steam volume = 0.05 m^{3}

Using the table of steam at 100 K pa

Steam density = 0.590 Kg/m^{3}

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Mass of vapor = S × V

m = 0.590 × 0.05

m = 0.0295 Kg

Solution:

a) The friction force is calculated

Friction force = In the given situation, the force need to stuck the piston.

= pressure inside the cylinder × piston area

= 100 × 10^{3}  × 0.1

f = 10000 N

b)  To calculate heat transfer.

Heat transfer = Heat needs drop temperatures 30°C.

Q=m c_{p} \ DT

Q = 0.0295 × 2.0267 × 10^{3} × 30

Q = 1.7936 KJ

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