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3 years ago

A hospital uses cobalt-60 in its radiotherapy treatments for cancer patients. Cobalt-60 has a half-life of 7

Mathematics

1 answer:

NeTakaya3 years ago

4 0

Answer:

We have 197 g of Co-60 after 18 months.

Step-by-step explanation:

We can use the decay equation.

$M_{f}=M_{i}e^{-\lambda t}$

Where:

M(f) and M(i) are the final and initial mass respectively
λ is the decay constant (ln(2)/t(1/2))
t(1/2) is the half-life of Co
t is the time at the final amount of m

$M_{f}=228e^{-\frac{ln(2)}{7} 1.5}$

$M_{f}=197\: g$

Therefore, we have 197 g of Co-60 after 18 months.

I hope it helps you!

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Strain-displacement relationship) Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loa

Anastasy [175]

Answer:

please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

$0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0$

The question also has 3 parts given as

Part a: Sketch the deformed shape for α=0.03, β=-0.01 .

Solution

As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.

the new points are calculated as follows

Point A(x=0,y=0)

Point A'(x+αx,y+βy)

Point A'(0+(0.03)(0),0+(-0.01)(0))

Point A'(0,0)

Point B(x=1,y=0)

Point B'(x+αx,y+βy)

Point B'(1+(0.03)(1),0+(-0.01)(0))

Point B'(1.03,0)

Point C(x=1,y=1)

Point C'(x+αx,y+βy)

Point C'(1+(0.03)(1),1+(-0.01)(1))

Point C'(1.03,0.99)

Point D(x=0,y=1)

Point D'(x+αx,y+βy)

Point D'(0+(0.03)(0),1+(-0.01)(1))

Point D'(0,0.99)

So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)

The plot is attached with the solution.

Part b: Calculate the six strain components.

Solution

Normal Strain Components

$\epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\$

Shear Strain Components

$\gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0$

Part c: Find the volume change

$\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\$

Also the change in volume is 0.0197

For the unit cube, the change in terms of strains is given as

$\Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the strain values are small second and higher order values are ignored so

$\Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the initial volume of cube is unitary so this result can be proved.

5 0

3 years ago

A system of inequalities is shown below.

marissa [1.9K]

Answer:

ccc

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Look at the diagram, how do you know that y is not directly proportional to x.

lesya [120]

Step-by-step explanation:

If y was directly proportional to x, the graph of y against x would pass through the origin (O).

Since the graph above does not show this, y is not directly proportional to x.

7 0

2 years ago

Original price: $32.00 percent of markup: 10%

densk [106]

Answer:

origional price of $32.

marked up 10%- $3.20

new price- $35.20

6 0

3 years ago

On a coordinate plane, parallelogram A B C D has points (3, 6), (6, 5), (5, 1), and (2, 2).

Vadim26 [7]

Answer:

13 square units

Step-by-step explanation:

We need to find the distance between the points using the

Pythagorean Theorem, a² + b² = c²

(3,6) and (6,5); 6 - 3 = 3 and 6 - 5 = 1

3² + 1² = c²

9 + 1 = c²

c = √10

(6,5) and (5,1); 6 - 5 = 1 and 5 - 1 = 4

1² + 4² = c²

1 + 16 = c²

c = √17

√17 + √10 = 13.038

3 0

2 years ago