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3 years ago

Examples of sources of physical entropy for secure random number generation include:________

Physics

1 answer:

Hatshy [7]3 years ago

5 0

Answer:

Noise in the environment

Electrons emitted via the photoelectric effect

Explanation:

Physical entropy can be seen as a disorderliness that can be observed in a physical system. The noise in a physical system can be used for secure number generation because of its randomness.

Noise is known to have no defined pattern and hence, can be used to creates secure random key generations.

Electrons emitted via the photoelectric effect: Electrons are emitted in a random order whenever a beam of light is incident on a photoelectric material. These electrons emitted can be used for random key generations

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At high noon, the sun delivers 1150 w to each square meter of a blacktop road. if the hot asphalt loses energy only by radiation

Pie

Stephen`s Law:
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
$T = \sqrt[4]{202.857 * 10 ^{8} }$
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K.

3 0

3 years ago

Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it

d1i1m1o1n [39]

Answer:

Q = 165.95 cm³ / s, 1) v = $\sqrt{0.55^2 + 19.6 y}$ , 2) v = 2.05 m / s,

3) d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

Q = v A

where Q is the flow rate, A is area and v is the velocity

the area of a circle is

A = π r²

radius and diameter are related

r = d / 2

substituting

A = π d²/4

Q = π/4 v d²

let's reduce the magnitudes

v = 0.55 m / s = 55 cm / s

let's calculate

Q = π/4 55 1.96²

Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

v = $\sqrt{0.55^2 + 2 \ 9.8\ y}$

v = $\sqrt{0.55^2 + 19.6 y}$

2) ask to calculate the velocity for y = 0.2 m

v = $\sqrt{0.55^2 + 19.6 \ 0.2}$

v = 2.05 m / s

3) We write the continuous equation for this point 2

Q = v₂ A₂

A₂ = Q / v₂

let us reduce to the same units of the SI system

Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

A₂ = 165.95 10⁻⁶ / 2.05

A₂ = 80,759 10⁻⁶ m²

area is

A₂ = π/4 d₂²

d₂ = $\sqrt{4 A_2 / \pi }$

d₂ = $\sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }$

d₂ = 10.14 10⁻³ m

d₂ = 1.014 cm

4 0

3 years ago

Choose what colors are absorbed when white light hits a red apple. (Pick all that apply.)

astra-53 [7]

A red apple absorbs all colors of visible light except red, so red light
is the only light left to bounce off of the apple toward our eyes.
(This is a big part of the reason that we call it a "red" apple.)

Here's how the various items on the list make out when they hit the apple:

Red . . . . . reflected
Orange . . absorbed
Yellow . . . absorbed
Green . . . absorbed
Blue . . . . absorbed
Violet . . . absorbed
Black . . . no light; not a color
White . . . has all colors in it

4 0

3 years ago

Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c

Mariulka [41]

Answer:

A. $v_{3}=12.17m/s$

B. $v_{car}=6.3m/s\\v_{truck}=-6.3m/s$

C. ΔK $=-4.13x10^3J$

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

$v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg$

A. Since the two vehicles become entangled the final mass is:

$m_{3}=102kg+103kg=205kg$

From linear momentum we got that:

$p_{1}=p_{2}$

$m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}$

$v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}$

$v_{3}=12.17m/s$

B. The change in velocity of both vehicles are:

For the car

$v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s$

For the truck

$v_{truck}=12.17m/s-18.5m/s=-6.3m/s$

C. The change in kinetic energy is:

ΔK= $K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})$

ΔK= $\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J$

ΔK $=-4.13x10^{3}J$

6 0

3 years ago

A mercury thermometer is used to measure the temperature of boiling water. Why?

Rama09 [41]

Answer:

It has very high density, so a small bulb of a thermometer can contain much mercury. Mercury remains liquid state over a quite wide range of temperature because it freezes at 39°C and boils at 357°C.

Explanation:

6 0

3 years ago