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svetoff [14.1K]

4 years ago

15

A uranium-238 atom can break up into a thorium-234 atom and a particle called an alpha particle, αα-4. The numbers indicate the

inertias of the atoms and the alpha particle in atomic mass units (1 amuamu = 1.66 × 10−27 kg). When an uranium atom initially at rest breaks up, the thorium atom is observed to recoil with an x component of velocity of -2.2 × 10^5 m/s.
How much of the uranium atom's internal energy is released in the breakup?

1 answer:

alexdok [17]4 years ago

3 0

Answer: E = 5.80*10^-13 J

Explanation:

Given

We use the law of conservation of momentum to solve this

Momentum before breakup = momentum after breakup

0 = m1v1 + m2v2

0 = 238m * -2.2*10^5 + 4m * v2

0 = -523.6m m/s + 4m * v2

v2 * 4m = 523.6m m/s

v2 = 523.6 m m/s / 4m

v2 = 130.9*10^5 m/s

v2 = 1.31*10^7 m/s

Using this speed in the energy equation, we have

E = 1/2m1v1² + 1/2m2v2²

E = 1/2 * (238 * 1.66*10^-27) * -2.2*10^5² + 1/2 * (4 * 1.66*10^-27) * 1.31*10^7²

E = [1/2 * 3.95*10^-25 * 4.84*10^10] + [1/2 * 6.64*10^-27 * 1.716*10^14]

E = (1/2 * 1.911*10^-14) + (1/2 * 1.139*10^-12)

E = 9.56*10^-15 + 5.7*10^-13

E = 5.80*10^-13 J

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Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

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4 years ago

A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force

Alexeev081 [22]

Answer:

This distance is measured from the center of the earth r = 3.4 10⁸ m

Explanation:

The equation for gravitational attraction force is

F = G m1 m2 / r²

Where g is the universal gravitation constant, m are the masses of the body and r is the distance between them, remember that this force is always attractive

Let's write the sum of force on the ship and place the condition that is balanced

F1 -F2 = 0

F1 = F2

Let's write this equation for our case

G m Me / r² = G m Mm / (r'.)²

The distance r is measured from the center of the earth and the distance r' is measured from the center of the moon,

r' = 3.85 10⁸ m

Let's simplify and calculate the distance

Me / r² = Mm / / (3.85 108- r)²

Me / Mm (3.85 108- r)² = r²

√ 81.4 (3.85 108 -r) = r

√ 81.4 3.85 108 = r (1 + √ 81.4)

34.74 108 = r (10.02)

r = 34.74 10⁸ / 10.2

r = 3.4 10⁸ m

This distance is measured from the center of the earth

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4 years ago

Mt. Everest is 29,028 feet high. How many miles is this?

Viefleur [7K]

Mount. everest is 5.499 miles

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4 years ago

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aniked [119]

Answer:

232

U

92

Explanation:

mass number=protons add neutrons

Atomic number= amount of protons

8 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago