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3 years ago

The waste products of a nuclear fission powerplant can best be described as

Physics

2 answers:

Nutka1998 [239]3 years ago

7 0

The answer is D. small in quantity and very dangerously radioactive.

OverLord2011 [107]3 years ago

4 0

A.

-> the products are stable atoms with a much lower atomic number. For example, Uranium 238 commonly breaks down into strontium and xenon, both of which are stable and non-radioactive.

They are also low in quantity because of the precision involved in breaking down a large atom via neutron capture means that few fission reactions take place.

Hope I helped :)

Source: I studied nuclear physics in year 11

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An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and

svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

$V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s$

velocity at the exit=20m/s

for entry

$V=\frac{5}{1} =5m/s$

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

$\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}$

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

$(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81 =p2\\P2=-37.5kPa$

the pressure at exit is -37.5kPa

7 0

3 years ago

A uniformly charged sphere has a total charge of 300uc and a radius of 8cm. Find the electric field density at A point 16cm from

s2008m [1.1K]

E = kQ 

(r + h)²

where,

k = 9 × 10^9Nm²C^-2

Q = total charge, 300uC = 300 × 10^ -6C

r = 8 × 10^ -2m

h = 16 × 10^ -2m

then,

E = 9e9 * 300e^-6 

(8e^-2 + 16e^-2)²

E = 4687500N/C

6 0

2 years ago

What is the frequency of radiation whose wavelength is 2.40 x 10-5 cm?

S_A_V [24]

To begin, the formula for finding frequency when wavelength is known is "f = c / w" when c is the constant velocity (3 * 10^8 m/s). To convert the wavelength into a common form (m/s), it will have to be multiplied by 10^-2. This leaves the equation as "f = 3.0 * 10^8 / (2.4 * 10^-5 * 10^-2), or 2.4 * 10^-7. This gives 1.25 * 10^15 m/s as the frequency.

7 0

3 years ago

An experimental tungsten light bulb filament has a length of 5 cm and a diameter of 0.074 cm. The filament is basically just a w

adell [148]

Answer:

power emitted is 1.75 W

Explanation:

given data

length l = 5 cm = 5 × $10^{-2}$ m

diameter d = 0.074 cm = 74 × $10^{-5}$ m

total filament emissivity = 0.300

temperature = 3068 K

to find out

power emitted

solution

we find first area that is π×d×L

area = π×d×L

area = π×74 × $10^{-5}$ ×5 × $10^{-2}$

area = 1162.3892 × $10^{-5}$ m²

so here power emitted is express as

power emitted = E × σ × area × (temperature)^4

put here all value

power emitted = 0.300× 5.67 × 1162.3892 × $10^{-5}$ × (3068)^4

power emitted = 1.75 W

5 0

3 years ago

The rules and expectations concerning correct or polite behavior is called _______________________.

aalyn [17]

The answer would be etiquette!

7 0

3 years ago