Neptunium. In the fall of 2002, scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. Neptunium-237 has a density of 19.5 g/cm3. What would be the radius of a sphere of this material that has a critical mass?

Answers

Answer 1

Answer:

To solve this problem it is necessary to apply the concepts related to density, such as the relationship between density and Volume.

The volume of a sphere can be expressed as

$V = (4)/(3) \pi r^3$

Here r is the radius of the sphere and V is the volume of Sphere

Using the expression of the density we know that

$\rho = (m)/(V) \rightarrow V = (m)/(\rho)$

The density is given as

$\rho = (19.5g/cm^3)((10^3kg/m^3)/(1g/cm^3))$

$\rho = 19.5*10^3kg/m^3$

Now replacing the mass given and the actual density we have that the volume is

$V = (60kg)/(19.5*10^3kg/m^3 )$

$V = 3.0769*10^(-3) m ^3$

The radius then is,

$V = (4)/(3) \pi r^3$

$r = \sqrt[3]{(3V)/(4\pi)}$

Replacing,

$r = \sqrt[3]{(3(3.0769*10^(-3)))/(4\pi)}$

The radius of a sphere made of this material that has a critical mass is 9.02 cm.

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1 pt each. Using the table above as a guide, complete the following conversions. Be sure to show your work to the side:
1. 5 cm = ________ mm
2. 83 cm = ________ m
3. 459 L = _______ ml
4. .378 Kg = ______ g
5. 45°F = ________ °C
6. 80°C = _________ °F

Answers

$5cm = 50mm$
$2.83cm = 0.0283m$
$3.459l = 3459ml$
$4.378kg = 4378g$
$5.45f = - 47.79c$
$6.80c = 44.24f$

With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 41 m

Answers

Answer:

The speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

Explanation:

Given;

maximum vertical height of the throw, H = 41 m

Apply the following kinematic equation;

V² = U² + 2gH

where;

V is the final speed with which the ball will rise to a maximum height

U is the initial speed of the ball = 0

g is acceleration due to gravity = 0

V² = U² + 2gH

V² = 0² + 2gH

V² = 2gH

V = √2gH

V = √(2 x 9.8 x 41)

V = 28.35 m/s

Therefore, the speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

Find the magnitude and direction of an electric field that exerts a 4.80×10−17N westward force on an electron. (b) What magnitude and direction force does this field exert on a proton?

Answers

a). The magnitude along with the direction of the electric field releasing westward force of $4.80$ × $10^(-17)$ N would be:

$3$ × $10^(-36)$ N/C is Eastward Direction

b). The magnitude along with the force of the direction that this field releases on proton would be:

$4.8$ × $10^(17) N$ in Eastward Direction

Electric Field

a). Given that,

Force $=$ $4.80$ × $10^(-17)$ N

As we know,

Force $=$ Charge × Electric Field

So,

∵ Electric Field $= Force/Charge$

$= 4.8$ × $10^(17)$ ) $/(1.6$ × $10^(-19)$ $)$

$= 3$ × $10^(36)N/C$

The direction of the field would be opposite i.e. Eastward direction due to the field carrying a -ve charge.

b). The magnitude carried by the force working on the proton would be the same with an opposite direction due to +ve charge.

∵ Force $=$ $4.80$ × $10^(-17)$ N in Eastward direction.

Learn more about "Magnitude" here:

brainly.com/question/9774180

Explanation:

(a) E = F/q

E = 4.8×10^-17/1.6×10^-19

E = 300 N/C

(b) same magnitude of electric field is exerted on proton

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equal amounts of energy are added to them. Assuming no melting or vaporization occurs, which of the following can be concluded about the final temperature TA of substance A and the final temperature TB of substance B?a) TA > TB
b) TA < TB
c) TA = TB
d) More information is needed

Answers

The final temperatures are such that TA > TB.

The specific heat capacity refers to the quantity of heat required to raise the temperature of 1 Kg of a body by 1K. The higher the specific heat capacity of a body, the higher the quantity of heat required to raise the temperature of the body and vice versa.

Hence, if the specific heat of substance A is greater than that of substance B and A and B are at the same initial temperature, when equal amounts of energy are added to them, the final temperature are such that TA > TB.

Learn more: brainly.com/question/1445383

Answer:

$m_A c_(pA) (T_(fA) -T) = m_B c_(pB) (T_(fB)- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

$c_(pA) > c_(pB)$

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

$T_(iA)= T_(iB)$

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

$Q = m c_p \Delta T$

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

$Q_A = Q_B$

And if we replace the formula for sensible heat we got:

$m_A c_(pA) \Delta T_A = m_B c_(pB) \Delta T_B$

And if we replace for the change of the temperature we got:

$m_A c_(pA) (T_(fA) -T_(iA)) = m_B c_(pB) (T_(fB)- T_(iB))$

And since $T_(iA)= T_(iB)= T$ we have this:

$m_A c_(pA) (T_(fA) -T) = m_B c_(pB) (T_(fB)- T)$

d) More information is needed

(The relation between the masses is not given)

Which of the following best represents stored potential energy?Air leaking from a flat tire
Stress built up in a rock fault
Heat given off by a forest fire
Water flowing through a hose

Answers

Answer:

Explanation:

stress built up on a rock fault

A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The block rests on a frictionless surface. A 5.30×10?2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.97 m/s and sticking.Part AHow far does the putty-block system compress the spring?

Answers

The distance the putty-block system compress the spring is 0.15 meter.

Given the following data:

Mass = 0.454 kg

Spring constant = 21.0 N/m.

Mass of putty = $5.30* 10^(-2)\;kg$

Speed = 8.97 m/s

To determine how far (distance) the putty-block system compress the spring:

First of all, we would solver for the initialmomentum of the putty.

$P_p = mass * velocity\n\nP_p = 5.30* 10^(-2)* 8.97\n\nP_p = 47.54 * 10^(-2) \;kgm/s$

Next, we would apply the law of conservation of momentum to find the final velocity of the putty-block system:

$P_p = (M_b + M_p)V\n\n47.54* 10^(-2) = (0.454 + 5.30* 10^(-2))V\n\n47.54* 10^(-2) = 0.507V\n\nV = (0.4754)/(0.507)$

Velocity, V = 0.94 m/s

To find the compression distance, we would apply the law of conservation of energy:

$U_E = K_E\n\n(1)/(2) kx^2 = (1)/(2) mv^2\n\nkx^2 =M_(bp)v^2\n\nx^2 = (M_(bp)v^2)/(k) \n\nx^2 = ((0.454 + 5.30* 10^(-2)) * 0.94^2)/(21)\n\nx^2 = ((0.507 * 0.8836))/(21)\n\nx^2 = ((0.4480))/(21)\n\nx=√(0.0213)$