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2 years ago

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Q 32.35: The isotope 235U decays by alpha emission with a half-life of 7.0 x 108 y. It also decays (rarely) by spontaneous fissi

on, and if the alpha decay did not occur, its half-life due to spontaneous fission alone would be 3.0 x 1017 y. At what rate do spontaneous fission decays occur in 1.0 g of 235U

1 answer:

Julli [10]2 years ago

5 0

Answer:

rate of fission =5.89*10^3 1\Year

Explanation:

we know that

rate of fission is given as $rate of fission = \frac{0.69}{(T_{1/2})_{fission}} *\frac{ Mass* Avogardo\ number}{Molar\ mass}$

$(T_{1/2})_{fission} = 3*10^{17} y$

mass = 1.0 g

avogardo number = 6.02*10^23

molar mass of isotopes 235U =235

Putting all value to get rate of emission

rate of fission $= \frac{0.69}{3*10^{17}} *\frac{1.0*6.02*10^{23}}{235}$

rate of fission =5.89*10^3 1\Year

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The wavelength of green light is 550 nm.

SIZIF [17.4K]

Answer:

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velocity of electron is 1323.88 m/s

momentum of the electron is 1.205 x 10⁻²⁷ kgm/s

(b) momentum of photon is 1.506 x 10⁻²⁷ kgm/s

velocity of electron is 1654.85 m/s

momentum of the electron is 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron

Explanation:

(a)

wavelength of green light, λ = 550 nm

momentum of photon is given by;

$p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{550*10^{-9}}\\\\p = 1.205 *10^{-27} \ kg.m/s$

velocity of electron is given by;

$P = \frac{h}{\lambda} \\\\mv = \frac{h}{\lambda}\\\\v = \frac{h}{m \lambda}\\\\v = \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(550*10^{-9})}\\\\v = 1323.88 \ m/s$

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1323.88)

p = 1.205 x 10⁻²⁷ kgm/s

(b)

wavelength of red light, λ = 440 nm

momentum of photon is given by;

$p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{440*10^{-9}}\\\\p = 1.506 *10^{-27} \ kg.m/s$

velocity of electron is given by;

$v = \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(440*10^{-9})}\\\\v = 1654.85 \ m/s$

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1654.85)

p = 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron.

7 0

2 years ago

Variations in the resistivity of blood can give valuable clues about changes in various properties of the blood. Suppose a medic

Elena-2011 [213]

Answer:

Answer:

1.1 x 10^9 ohm metre

Explanation:

diameter = 1.5 mm

length, l = 5 cm

Potential difference, V = 9 V

current, i = 230 micro Ampere = 230 x 10^-6 A

radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m

Let the resistivity is ρ.

Area of crossection

A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2

Use Ohm's law to find the value of resistance

V = i x R

9 = 230 x 10^-6 x R

R = 39130.4 ohm

Use the formula for the resistance

$R=\rho \frac{l}{A}$

$\rho =\frac{RA}{l}$

$\rho =\frac{39130.4\times 0.05}{1.766\times 10^{-6}}$

ρ = 1.1 x 10^9 ohm metre

Explanation:

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2 years ago

What is the mass of an object that weighs 686N on Earth?

Oxana [17]

Answer:

70 kg is the mass of the object

Explanation:

This question can be solved with this simple formula:

Weight force = mass . gravity

686 N = mass . 9.8 m/s²

686 N / 9.8 m/s² = mass → 70 kg

Note → 1N = 1 kg . m / s²

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2 years ago

During a race the dirt bike was observed to leap up off the small hill at A at an angle of 60^o with the horizontal. If the poin

const2013 [10]

Answer:

Velocity is equal to $27.3$ feet per second and time is equal to $1.4668$ seconds

Explanation:

Given

The horizontal distance traveled by dirt bike before landing $= 20$ feet

Angle of flight $= 60$ degree

As we know that Horizontal distance (H) is equal to

$= H_0 + V_0 * t\\$

Where $H_0$ is the initial horizontal distance

$V_0$ is the velocity with which the bike is travelling in horizontal direction

and $t$ is the time in seconds

Substituting the given values, we get -

$H = H_0 + v*t\\20 = 0 + v * cos \theta * t\\20 = v * cos 60 * t\\t = \frac{20}{v * cos 60} \\t = \frac{40}{v}$

Now distance traveled in vertical direction is equal to

$Y = Y_0 + v_0 * t + \frac{1}{2} a * t^2$

here acceleration will be equal to acceleration due to gravity which is equal to $- 32.2 \frac{ft}{s^2}$ . It is negative as its is acting in upward direction

Thus,

$Y = 0 + v * sin 60 + \frac{1}{2} * (-32.2) * (\frac{40}{v} )^2\\0 = 0 + 0.866 v + \frac{-25760}{v^2} \\0.866 v = \frac{-25760}{v^2}\\v^3 = \frac{-25760}{0.866} \\v = 27.3$

Velocity is equal to $27.3$ feet per second and time is equal to $1.4668$ seconds

8 0

2 years ago

A thin, uniform rod is hinged at its midpoint. To begin with, one-half of the rod is bent upward and is perpendicular to the oth

Dima020 [189]

Answer:

$\omega_1 = 6.06 rad/s$

Explanation:

Initially rod is bent into L shaped at mid point

So we will have rotational inertia of the rod given as

$I = \frac{mL^2}{3} + (\frac{mL^2}{12} + m(L^2 + \frac{L^2}{4}))$

$I = \frac{5mL^2}{3}$

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then we will have

$I_2= \frac{(2m)(2L)^2}{3} = \frac{8mL^2}{3}$

now from angular momentum conservation we have

$I \omega_o = I_1 \omega_1$

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$\omega_1 = 6.06 rad/s$

7 0

2 years ago

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