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Kazeer [188]
3 years ago
12

U-235 releases an average of 2.5 neutrons per fission, while Pu-239 releases an average of 2.7 neutrons per fission. Which of th

ese elements might you therefore expect to have the smaller critical mass
Physics
1 answer:
Komok [63]3 years ago
5 0

Answer:

Pu - 239 have the smaller critical mass.

Explanation:

Critical mass is the smallest amount of certain element of mass that is needed to achieve a nuclear chain reaction early . Since Pu - 239 releases an average of 2.7 neutrons per fission as compared to U - 235 that releases 2.5 neutrons per fission. So, Pu - 239 has smaller critical mass, because Pu - 239 has a higher probability for fission and produces a large no. of neutrons per fission event. Infact of all the basic nuclear fuels, Pu - 239 has smallest critical mass. Critical mass depends on the nuclear properties of elements undergoing fission reaction. Hence, as Pu - 239 produces large no. of neutrons per fission than U - 235 and Pu - 239 has smaller critical mass.

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It became cool because many people who likes ⚽️ soccer felt it atrcting and then they spreaded the world cup out and then people started.likeing it slot
5 0
3 years ago
A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thic
Mazyrski [523]

Answer:

The total resistance of the wire is = 1.917\times10^{-3}

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, R_{Total}

\frac{1}{R_{Total}}  =\frac{1}{R_{cu}  }+\frac{1}{R_{al}}

Parameters given:

Length of wire = 1 m

Cross sectional area of copper A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3}  )^{2} =3.142\times10^{-6} m^{2}

Cross sectional area of aluminium wire  

A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3}  )^{2}-(1\times 10^{-3}  )^{2}] =9.42\times10^{-6} m^{2}\\

Resistivity of copper \rho _{cu}=1.7\times 10^{-8}  \Omega .m

Resistivity of Aluminium \rho _{al}=2.8\times 10^{-8}  \Omega .m

Resistance of copper R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} }  =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega

Resistance of aluminium R_{al}= \frac{\rho_{al} \times l}{A_{al} }  =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega

The total resistance of the wire can be obtained as follows;

\frac{1}{R_{Total}}  =\frac{1}{5.41\times10^{-3}  }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}

R_{Total}= 1.917\times 10^{-3}\Omega

∴ The total resistance of the wire = 1.917\times 10^{-3}\Omega

4 0
3 years ago
Read 2 more answers
Due Sun 06/09/2019 11:59 pm Skip Navigation Questions correct Q 1 [1/1] correct Q 2 [1/1] correct Q 3 [1/1] untried Q 4 (0/1) un
astraxan [27]

Due Sun 06/09/2019 11:59 pm <u><em>(you're already more than a week late)</em></u> Skip Navigation Questions, correct Q 1 [1/1], correct Q 2 [1/1], correct Q 3 [1/1], untried Q 4 (0/1), untried Q 5 (0/1), untried Q 6 (0/1), untried Q 7 (0/1), untried Q 8 (0/1), untried Q 9 (0/1), untried Q 10 (0/1), Grade: 3/10, Print Version, <u><em>Start of Questions</em></u>:

Questions Two cyclists, 42 miles apart, start riding toward each other at the same time. One cycles 2 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of the faster cyclist?

<u><em>Start of Answer:</em></u>

-- When they started, they were 42 miles apart.  When they met, 2 hours later, they were no miles apart.  The distance between them shrank at the rate of 21 miles per hour.  Since they rode directly toward each other, the sum of their individual speeds must have been 21 miles per hour.

-- One biker was two times the speed of the other.  Slower biker: 1 time.  Faster biker: 2 times.  Sum of their speeds:  3 times = 21 miles per hour.  Each 'time' = 7 miles per hour.

-- Slower cycler = 1 time = 7 mi/hr

<em>Faster cycler = 2 times = 14 mi/hr</em>

5 0
3 years ago
You are approaching a helicopter on a helipad in an ambulance. The helicopter is powered down and the blades are not spinning. A
Maksim231197 [3]

After notifying the PIC of your arrival, it is best to approach the sides of the helicopter in this scenario.

<h3>What is an Ambulance?</h3>

This is a vehicle which contains medical equipments and are involved in the transportation of patients to healthcare facilities.

It is best to approach the sides of the helicopter and not tail rotor end to ensure visibility and safety.

Read more about Ambulance here brainly.com/question/16612316

#SPJ1

7 0
2 years ago
Two gliders are on a frictionless, level air track. Both gliders are free to move. Initially, glider A moves to the right and gl
Yuliya22 [10]

Answer:

The change in momentum of both objects is the same but in opposite direction.

Explanation:

Hi there!

The momentum of the system is calculated as the sum of the momentums of each glider. The momentum of the system is conserved if no external force is acting on the objects (as in this case). That means that the initial momentum of the system is equal to the final momentum of the system.

The momentum of each glider is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass of the glider.

v = velocity.

The momentum of the system for glider A and B can be calculated as follows:

initial momentum = mA · vA + mB · vB

Where:

mA and vA = mass and velocity of glider A

mB and vB = mass and velocity of glider B

Initially, glider B is at rest so that vB = 0. Then, the initial momentum of the system is:

initial momentum = mA · vA

The final momentum of the system is calculated as follows:

final momentum = mA · vA´ + mB · vB´

Where vA´ and vB´ are the final velocities of glider A and B respectively.

We know that mB = 4mA and that vA´ is negative. The the final momentum will be:

final momentum = -mA · vA´ + 4mA · vB´

Since initial momentum = final momentum:

mA · vA = -mA · vA´ + 4mA · vB´

mA · vA + mA · vA´ = 4mA · vB´

<u>vA + vA´ = 4 vB´</u>

<u />

The change in momentum of glider A (ΔpA) is calculated as follows:

ΔpA = final momentum - initial momentum

ΔpA =  -mA · vA´ - mA · vA = -mA (vA + vA´) = -4mA · vB´

The change in momentum of glider B (ΔpB) is calculated as follows:

ΔpB = final momentum - initial momentum

ΔpB = 4mA · vB´ - 0 = 4mA · vB´

Then, the change in momentum of both objects is the same but in opposite direction. That´s why the momentum is conserved.

4 0
2 years ago
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