Ferroconcrete is reinforced concrete that combines concrete and _______

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

Wittaler [7]

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

4 0

3 years ago

What size resistor would produce a current flow of 5 Amps with a battery voltage of 12.6 volts

Debora [2.8K]

Answer:

resistance = 2.52 ohms

Explanation:

from the formula

V =IR

Voltage = (current)(resistance)

Resistance =

R=

R= 2.52 ohms

5 0

2 years ago

A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.

denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is $8.621 \times 10^{-4} in/in$

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

$\frac{M}{I}=\frac{F}{y}$

Here

M is the moment which is to be calculated
I is the moment of inertia given as

$I=\frac{bd^3}{12}$

Here

b is the breath given as 0.75"
d is the depth which is given as 8"

$I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4$

y is given as

$y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\$

Force is 50 ksi

$\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in$

The yield moment is 400 k.in.

Part b:

The strain is given as

$Strain=\frac{Stress}{Elastic Modulus}$

The stress at the station 2" down from the top is estimated by ratio of triangles as

$F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi$

Now the steel has the elastic modulus of E=29000 ksi

$Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in$

So the strain is $8.621 \times 10^{-4} in/in$

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

$shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi$

The plastic moment is 600 ksi.

3 0

3 years ago

Was rachel and ross on a break

yaroslaw [1]

Answer:

yes they were on a break

Explanation:

3 0

3 years ago

A hypothetical metal has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.413 nm, 0.665 nm, and 0.87

Inessa [10]

Answer:

atomic radius R = 0.157 nm

metal atomic weight = 72.27 g/mol

Explanation:

given data

parameters a = 0.413 nm

parameters b = 0.665 nm

parameters c = 0.876 nm

atomic packing factor = 0.536

density = 3.99 g/cm³

to find out

atomic radius and atomic weight

solution

we apply here atomic packing factor (x) that is

atomic packing factor (x) = $\frac{volume(sphere)}{volume(unit\ cell)}$ ..................1

put here value we get

atomic packing factor = $\frac{8*(4/3)*\pi R^3}{3*a*b*c}$

R = $(\frac{3(x)(abc)}{32\pi })^{1/3}$

R = $(\frac{3(0.536)(0.413*0.665*0.876)}{32\pi })^{1/3}$

atomic radius R = 0.157 nm

and

now we get here metal atomic weight that is

metal atomic weight = $\frac{\rho (abc)(N_A)}{no\ of\ atom}$ ....................2

metal atomic weight = $\frac{3.99 (0.413*0.665*0.876)(6.023*10^{23})}{8}$

metal atomic weight = 72.27 g/mol

6 0

3 years ago

Ferroconcrete is reinforced concrete that combines concrete and ________. A. Lead c. Copper b. Iron d. Aluminum.