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Valentin [98]
2 years ago
6

Flip-flops are normally used for all of the following applications, except ________. logic gates data storage frequency division

counting
Engineering
1 answer:
SpyIntel [72]2 years ago
6 0

Flip-flops are normally used for all of the following applications, except  logic gates.

<h3>What are Flip flops?</h3>

Flip flops are known to be tools that are used for counting. They come in different ranges.

Note that Flip flops are one that can be seen on counters, storage registers, and others and as such, Flip-flops are normally used for all of the following applications, except  logic gates.

Learn more about  Flip flops from

brainly.com/question/4237777

#SPJ1

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A closed system consisting of 4 lb of a gas undergoes a process during which the relation between pressure and volume is pVn 5 c
gayaneshka [121]

Answer:

V1=5<u>ft3</u>

<u>V2=2ft3</u>

n=1.377

Explanation:

PART A:

the volume of each state is obtained by multiplying the mass by the specific volume in each state

V=volume

v=especific volume

m=mass

V=mv

state 1

V1=m.v1

V1=4lb*1.25ft3/lb=5<u>ft3</u>

state 2

V2=m.v2

V2=4lb*0.5ft3/lb=  <u> 2ft3</u>

PART B:

since the PV ^ n is constant we can equal the equations of state 1 and state 2

P1V1^n=P2V2^n

P1/P2=(V2/V1)^n

ln(P1/P2)=n . ln (V2/V1)

n=ln(P1/P2)/ ln (V2/V1)

n=ln(15/53)/ ln (2/5)

n=1.377

3 0
3 years ago
List and describe three classifications of burns to the body.
DiKsa [7]

AnswerWhat Are the Classifications of Burns? Burns are classified as first-, second-, or third-degree, depending on how deep and severe they penetrate the skin's surface. First-degree burns affect only the epidermis, or outer layer of skin. The burn site is red, painful, dry, and with no blisters.

Explanation:

8 0
3 years ago
Read 2 more answers
Air enters a tank through an area of 0.2 ft2 with a velocity of 15 ft/s and a density of 0.03 slug/ft3. Air leaves with a veloci
Mademuasel [1]

Answer:

please find attached.

Explanation:

4 0
3 years ago
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
The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam
lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy

T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0
3 years ago
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