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3 years ago

A Decay Chain

Physics

1 answer:

Strike441 [17]3 years ago

7 0

Answer : The energy released in first step of thorium-232 decay chain is $7.974\times 10^{-13}J$

Explanation :

First we have to calculate the mass defect $(\Delta m)$ .

The balanced reaction is,

$^{232}Th\rightarrow ^{228}Ra+^{4}He$

Mass defect = Sum of mass of product - sum of mass of reactants

$\Delta m=(\text{Mass of Ra}+\text{Mass of He})-(\text{Mass of Th})$

$\Delta m=(228.0301069+4.002602)-(232.038054)=5.34\times 10^{-3}amu=8.86\times 10^{-30}kg$

conversion used : $(1amu=1.66\times 10^{-27}kg)$

Now we have to calculate the energy released.

$Energy=\Delta m\times (c)^2$

$Energy=(8.86\times 10^{-30}kg)\times (3\times 10^8m/s)^2$

$Energy=7.974\times 10^{-13}J$

The energy released is $7.974\times 10^{-13}J$

Therefore, the energy released in first step of thorium-232 decay chain is $7.974\times 10^{-13}J$

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A 62.2 kg ice skater moving to the right with a velocity of 2.66 m/s throws a 0.145 kg snowball to the right with a velocity of

Travka [436]

Answer:

Velocity of skater after throwing the snowball is 2.57 m/s

Explanation:

Given :

Mass of skater, M = 62.2 kg

Mass of snowball, m = 0.145 kg

Velocity of snowball relative to ground, v = 39.3 m/s

Consider v₁ be the velocity of skater after throwing the snowball.

According to the problem, initially the velocity of skater and snowball is same. So,

Velocity of skater before throwing snowball, u = 2.66 m/s

Applying conservation of momentum,

Momentum before throwing snowball = Momentum after throwing snowball

(M + m) u = Mv₁ + mv

$v_{1}=\frac{(M+m)u-mv}{M}$

Substitute the suitable values in the above equation.

$v_{1}=\frac{(62.2+0.145)2.66-0.145\times39.3}{62.2}$

v₁ = 2.57 m/s

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4 years ago

the lever in the figure below accomplishes 5.0 newtons-meters of work raising a 25 newton weight. How far,in meters is the weigh

miv72 [106K]

The simple definition of work is the product of the force applied to the object and the displacement:

W = Fd

W is the work, F is the applied force, and d is the displacement.

Givens:

W = 5.0N-m

F = 25N (the applied force must be at least equal to the weight to lift it)

Plug in the given values and solve for d:

5.0 = 25d

d = 0.20m

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3 years ago

Question 1

Vesna [10]

Answer:

1) g = 4π² / m, 3) xaxis the length of the pendulums and the y axis the period squared

Explanation:

a) students can approximate this system to a simple pendulum, in this case the angular velocity is

w = √ g / l

angular velocity, frequency and period are related

w = 2π f = 2π / T

we substitute

T = 2π√ l / g

with this equation they can determine the value of the acceleration of gravity, for this they measure the period for various lengths of the pendulum and graph

T² = 4π² l / g

We graph T² vs l

where this is the equation of a line if the independent variable is y = T² and x = l

y = (4π² / g) l

so the slope is

m = 4π² / g

clearing

g = 4π² / m

where the slope can be found with the values of the line not the experimental values.

2) to carry out the experiment, or the thread is attached to the sphere, the length of the pendulum that goes from the pivot point to the center of the sphere is measured with a tape measure and a small finished angle is turned or less than 10th is released, it is good to wait for the first oscillation to walk, the time of a determined number of oscillations is generally measured 10 or 20, the period is calculated

T = t / n

a table of T² against the length is made and it is plotted with the length in the ax ax, we look for the slope and hence the acceleration of gravity

3) on the independent x-axis, the controlled variable must be plotted, which is the length of the pendulums, and on the y-axis, the dependent variable is the period squared

4) of the equation of the line

m = 4pi2 / g

where it ends up reaching the floor

g = 4pi2 / m

5) when the spring is cut, the sphere remains under the effect of gravity acceleration, the harmonic movement disappears and the sphere is in a vertical movement

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4 years ago

Resistance and resistivity are related by a: 1.factor that is dependent on the length of the material 2.factor that is dependent

luda_lava [24]

Answer:

1. factor that is dependent on the length of the material.

2. factor that is dependent on the area of the material

Explanation:

Resistance R is directly proportional to its length L of an object and it is inversely proportional to the crossectional area of that object.

Mathematically, this is represented as:

a) Resistance R is directly proportional to its length L of an object

R ∝ L ......... Equation 1

b) Resistance R is inversely proportional to the crossectional area (A) of an object

R ∝ 1/A ............ Equation 2

Combining the two equations together, we have:

Resistance R is directly proportional to its length L of an object

R ∝ L/A .......... Equation 3

Resistivity of a material is the natural or inherent or intrinsic property that a material has. The Resistivity of a material is not dependent on the length or crossectional area of such material. It is only temperature dependent.

The resistance of a material is also dependent on the resistivity of the material. This means the resistance of a material is directly proportional to the resistivity of the material.

R ∝ ρ........Equation 4

Combining Equation 4 and 3 together, we have:

R = ρL/A

Where:

R = Resistance

ρ = Resistivity

L = Length of the material

A = Crossectional area of the material

Resistance and resistivity are related by a factor that is dependent on the length of the material a factor that is dependent on the area of the material

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4 years ago

If Maria lifts a mass of 10kg into a table of 1.25m high. How much gpe has the mass gained

SCORPION-xisa [38]

Answer:

gravitational potential energy of 10 kg mass is 122.625 J

Explanation:

Gravitational Potential energy gain is given as

$U = mgh$