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2 years ago

6

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the rea

ding of 400. counts has diminished to 100. counts after 66.7 minutes , what is the half-life of this substance? Express your answer with the appropriate units.

1 answer:

vovangra [49]2 years ago

3 0

Answer: 33.35 minutes

Explanation:

A(t) = A(o) *(.5)^[t/(t1/2)]....equ1

Where

A(t) = geiger count after time t = 100

A(o) = initial geiger count = 400

(t1/2) = the half life of decay

t = time between geiger count = 66.7 minutes

Sub into equ 1

100=400(.5)^[66.7/(t1/2)

Equ becomes

.25= (.5)^[66.7/(t1/2)]

Take log of both sides

Log 0.25 = [66.7/(t1/2)] * log 0.5

66.7/(t1/2) = 2

(t1/2) = (66.7/2 ) = 33.35 minutes

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Answer:

c. V2 equals V1

Explanation:

We can answer this question by using the continuity equation, which states that:

$A_1 v_1 = A_2 v_2$ (1)

where

A1 is the cross-sectional area in the first section of the pipe

A2 is the cross-sectional area in the second section of the pipe

v1 is the velocity of the fluid in the first section of the pipe

v2 is the velocity of the fluid in the second section of the pipe

In this problem, we are told that the pipe has a uniform cross sectional area, so:

A1 = A2

As a consequence, according to eq.(1), this means that

v1 = v2

so, the velocity of the fluid in the pipe does not change.

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Polymer ropes and lines for use on water are often designed to float, to aid in their retrieval and to avoid applying a downward

andrew-mc [135]

Answer:

We choose PTFE

Explanation:

Attached are the modulus density and modulus strength chart.

Due to its young modulus, the density is near 0.5 GPa, as seen in the chart and support water gliding. The PTFE density is between 1 and 10 Mg / cubic meter (see module and chart of density), and the resistance is between 10 and 100 Mpa (see module and chart of strength). Therefore, the finest ploymer will be PTFE that meets the requirements.

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3 years ago

In a tensile test on a steel specimen, true strain is 0.171 at a stress of 263.8 MPa. When true stress is 346.2 MPa, true strain

swat32

Answer:n=0.973

Explanation:

Given

When True strain $\left ( \epsilon _T_1\right )=0.171$

at $\sigma _1=263.8 MPa$

When True stress $\left ( \sigma _2\right )$ =346.2 MPa

true strain $\left ( \epsilon _T_2\right )$ =0.226

We know

$\sigma =k\epsilon ^n$

where $\sigma$ =True stress

$\epsilon$ =true strain

n=strain hardening exponent

k=constant

Substituting value

$263.8=k\left ( 0.171\right )^n------1$

$346.2=k\left ( 0.226\right )^n-----2$

Divide 1 & 2 to get

$\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n$

$1.312=\left ( 1.3216\right )^n$

Taking Log both side

$ln\left ( 1.312\right )=nln\left ( 1.3216\right )$

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2 years ago

A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of

larisa86 [58]

Answer:

The field strength needed is 0.625 T

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Given;

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peak emf = 24 V

The peak emf is given by;

emf₀ = NABω

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PV = m $\bar{R}$ T (1)

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V = volume

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$\bar{R} = \frac{R}{M}$

R= Rydberg's constant = 8.314 KJ/K

M = molar mass of water = 18 g/ mol

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$V = \frac{m\bar{R}T}{P}$

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m' = 2000 - 100 = 1900 kg

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At equilibrium, volume remain same

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P'V = m' $\bar{R}$ T'

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2 years ago