Describe how analog electronics is applied in using the photocopier or old land-line

Which term describes the action of a qubit that moves from superposition to 1 or 0 after measurement?

Mariana [72]

The term that describes the action of a qubit that moves from superposition to 1 or 0 after measurement is Collapse.

<h3>What is collapse?</h3>

Collapse is the process that lead to the movement of qubit from a a state of superposition to 1 or 0 after measurements which make it to remain in that state.

Therefore, The term that describes the action of a qubit that moves from superposition to 1 or 0 after measurement is Collapse.

Learn more about collapse below.

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4 0

2 years ago

A horizontal curve on a two-lane highway (10-ft lanes) is designed for 50 mi/h with a 6% superelevation. The central angle of th

fenix001 [56]

Answer:

The PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

Explanation:

From table 3.5 of Traffic Engineering by Mannering

R_v=835

R=835+(10ft/2)= 840 ft.

Now T is given as

T=R tan(Δ/2)

Here Δ is the central angle of curve given as 35°

So

T=R tan(Δ/2)

T=840 x tan(35/2)

T=840 x tan(17.5)

T=264.85

Now

STA PC=482+72-(2+64.85)=480+07.15

Also L is given as

L=(π/180)RΔ

Here R is the radius calculated as 840 ft, Δ is the angle given as 35°.

L=(π/180)RΔ

L=(π/180)x840 x35

L=512.87 ft

STA PT=480+07.15+5+12.87=485+20.02

Now Ms is the minimum distance which is given as

$M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\$

Here R_v is given as 835

SSD for 50 mi/hr is given as 425 ft from table 3.1 of Traffic Engineering by Mannering

So Ms is

$M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\M_s=835(1-cos(\frac{90 \times 425}{\pi 835}))\\M_s=26.92 ft$

Now for the clearance from the inside lane

Ms=Ms-lane length

Ms=26.92-5= 21.92 ft.

So the PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

6 0

3 years ago

A 25 kg iron block initially at 280°C is quenched in an insulated tank that contains 100 kg of water at 18°C. Assuming the water

defon

Answer: water at 18°C, w cw = 4.18 kJ/kg∙°C; at 27oC, Cfe = 0.554 kJ/kg∙°C. (Note that we could also have the following units for the heat capacities: kJ/kg∙K.) Assuming that the heat capacities are constant we have the following

The final temperature of the iron and the water will be the same: Tfe,2 = Tw,2 = T2. Substituting T2 for Tfe,2 and Tw,2, and solving for T2 gives the following result for the final temperature.

T2= ((Mw. Cw.T1w) + (Mfe.Cfe. T1fe))/((Mw.Cw) + (Mfe.Cfe))...equ 1

Where Me= mass of water= 100kg, Mfe =mass of iron = 25kg, T1fe = temp of iron before= 280°c,

Using substitution

T2= ((100*4.18*18) + (25*0.554*280))/ ((100*4.18) + (25*.554))

T2 = 26.4°c

So determine total entropy change

DStot = DSw + DSfe ...equat3

DStot = final entropy, DSw = entropy of water at T2, DSfe = final entropy of iron at T2 where,

DS = M.C. lin(T2/T1)...equ 5 temp is in Kelvin.

DSw = 100*4.18*lin(299.4/291) = 11.895

DSfe = 25*.554*lin(299.4/553) = -8.498.

Substituting answers into equa3

DStot = 11.895 - 8.498 = 3.397kj/kg*Kelvin

Explanation: the explanation is in the answers above..

3 0

3 years ago

Write multiple if statements:

lora16 [44]

Zrizorzlzfxxxgoxxxxpgxtoxxxhxuxyf

3 0

3 years ago

A steel rotating-beam test specimen has an ultimate strength Sut of 1600 MPa. Estimate the life (N) of the specimen if it is tes

ziro4ka [17]

Answer:

the life (N) of the specimen is 46400 cycles

Explanation:

given data

ultimate strength Su = 1600 MPa

stress amplitude σa = 900 MPa

to find out

life (N) of the specimen

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 1600

Se = 800 Mpa

and we know

Se for steel is 700 Mpa for Su ≥ 1400 Mpa

so we take endurance limit Se is = 700 Mpa

and strength of friction f = 0.77 for 232 ksi

because for Se 0.5 Su at $10^{6}$ cycle = (1600 × 0.145 ksi ) = 232

so here coefficient value (a) will be

a = $\frac{(f*Su)^2}{Se}$

a = $\frac{(0.77*1600)^2}{700}$

a = 2168.3 Mpa

so

coefficient value (b) will be

a = - $\frac{1}{3}$ log $\frac{(f*Su)}{Se}$

b = - $\frac{1}{3}$ log $\frac{(0.77*1600)}{700}$

b = -0.0818

so no of cycle N is

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value

N = $(\frac{ 900}{2168.3})^{1/-0.0818}$

N = 46400

the life (N) of the specimen is 46400 cycles

5 0

3 years ago