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4 years ago

What is the energy (in MeV) released in the alpha decay of 231Pa?

Physics

1 answer:

vlabodo [156]4 years ago

4 0

Answer: 3729.87MeV

Explanation:

²³¹₉₁Pa → ²²⁷₈₉Ac + ⁴₂He

Molar mass of of He = 4.0

Number of moles = 1

Na (Avogadro's number) = 6.023*10²³ atoms

Note : 1 mole = molarmass = Avogadro's number

Molar mass = 4.000 * 1 = 4g = 4*10^-3 kg

Mass = (number of moles * molar mass ) / Avogadro's number.

Mass = 4.0*10^-3 / 6.023*10^-23

Mass = 6.64 * 10^-27 kg

from Einstein's theory of relativity, energy E and mass m are interconvertible.

∇E = ∇mc²

E = (6.64*10^-27) * (3.0*10⁸)²

E = 5.98*10^-10J

1eV = 1.602*10-19J

XeV = 5.98*10-19J

X = (5.98*10^-10 * 1) / 1.602*10-19

X = 37328333958eV

X = 3729.87MeV

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A sample of Bismuth-212 has a mass of 2.64 grams (g) after 121 seconds (s). What was the initial mass of the sample if Bismuth-2

uysha [10]

The mass of a radioactive element at time t is given by
$m(t) = m_0 ( \frac{1}{2} )^{ \frac{t}{t_{1/2}} }$
where $m_0$ is the mass at time zero, while $t_{1/2}$ is the half-life of the element.

In our problem, $m(t)=2.64 g$ , t=121.0 s and $t_{1/2}=60.5 s$ , so we can find the initial mass $m_0$ :
$m_0= \frac{m(t)}{ (\frac{1}{2})^{t/t_{1/2}} } = \frac{2.64 g}{( \frac{1}{2} )^{121/60.5}} =4 \cdot 2.64 g=10.56 g$

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3 years ago

Someone is trying to balance a (110cm) plank with certain forces.

Mashutka [201]

Answer:

a. i. 30 Nm ii. This moment is a clockwise positive moment.

b. i. 15 Nm ii, This moment is a counter-clockwise negative moment.

c. i. The plank will not balance. ii. The plank would tip up.

d. 150 N

Explanation:

a) Calculate the moment of the 60N force (about O), then name its type.

i. Calculate the moment of the 60N force (about O)

Since moment = Force × perpendicular distance from point of moment ,

M = Fd

Since F = 60 N and d = 50 cm = 0.5 m

M = 60 N × 0.5 m = 30 Nm

ii. Then name its type.

This moment is a clockwise positive moment.

b) Calculate the moment of the 30N force (about O), then name its type.

i. Calculate the moment of the 30N force (about O),

Since moment = Force × perpendicular distance from point of moment ,

M' = F'd'

Since F' = 30 N and d' = 50 cm = 0.5 m

M' = 30 N × 0.5 m = 15 Nm

ii. Then name its type.

This moment is a counter-clockwise negative moment.

c) Will the plank balance? If not, which way will it tip?

i. Will the plank balance?

The plank will balance if the net moment on it is zero

So net moment, M' = positive moment - negative moment = M - M' = 30 Nm - 15 Nm = 15 Nm

Since the net moment on the plank is M" = 15 Nm ≠ 0,<u>the plank will not balance.</u>

ii Which way will it tip?

The plank would tip in the direction of the greater moment since the net moment is positive. <u>This moment tilts the plank in a clockwise direction, so the plank would tip up.</u>

d) What extra force would be needed at (B) to balance the plank?

The extra force must balance the net moment,

So M" = F"d" where F" = force and d" = distance of force from O = 10 cm = 0.10 m

F" = M"/d"

= 15 Nm/0.10 m

= 150 N

4 0

3 years ago

The unit light-year is a measure of

stiv31 [10]

A light-year is a unit of distance.

The definition of a light-year is the distance light travels in one year.

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4 years ago

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The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$