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myrzilka [38]
3 years ago
6

1) A substance has a half life of 20 years. what percentage would be left after 40 years?

Physics
1 answer:
Umnica [9.8K]3 years ago
5 0
1) The half-life is the time required for a substance to reduce to half its initial value. In formulas:
\frac{m(t)}{m_0}  = ( \frac{1}{2} )^{t/t_{1/2}} (1)
where
m(t) is the amount of substance left at time t
m0 is the initial mass
t_{1/2} is the half-life

In this problem, the half-life of the substance is 20 years:
t_{1/2} = 20 y
therefore, the fraction of sample left after t=40 years will be
\frac{m(t)}{m_0}=( \frac{1}{2})^ \frac{40 y}{20 y}  = ( \frac{1}{2})^2 =  \frac{1}{4}

So, only 1/4 of the original sample will be left, which corresponds to 25%.

2) We can use again formula (1), by re-arranging it:
m_0 =  \frac{m(t)} {( \frac{1}{2} )^{ \frac{t}{t_{1/2} }}}
If we use m(t)=10 g (mass of uranium left at time t), and t=4 t_{1/2} (the time is equal to 4 half lifes), we get
m_0 =  \frac{10 g}{ (\frac{1}{2})^4 } =16 \cdot 10 g = 160 g
So, the initial sample of uranium was 160 g.
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vovangra [49]

Answer:

c=71.4m/s

\theta=8.54\textdegree

Explanation:

From the question we are told that

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Generally Resolving vector mathematically

  sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6

Generally the equation Pythagoras theorem is given mathematically by

c^2=a^2+b^2

c^2=10.6^2 +(10.6+60)^2

c=\sqrt{10.6^2 +(10.6+60)^2}

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c=71.4m/s

b)Resultant direction

Generally the equation for solving Resultant direction

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Therefore

\theta=tan^-1(\frac{10.6}{70.6})

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3 years ago
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Answer:

(c) no different than on a low-pressure day.

Explanation:

The force acting on the ship when it floats in water is the buoyant force. According to the Archimedes' principle: The magnitude of buoyant force acting on the body of the object is equal to the volume displaced by the object.

Thus, Buoyant forces are a volume phenomenon and is determined by the volume of the fluid displaced.  

<u>Whether it is a high pressure day or a low pressure day, the level of the floating ship is unaffected because the increased or decreased pressure at the all the points of the water and the ship and there will be no change in the volume of the water displaced by the ship.</u>

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Hope this helps!

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An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s 1045 rad/s ). If a particular disk is
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Answer:

The magnitude of the average angular acceleration of the disk is 4139.74\ rad/s^2.

Explanation:

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Time, t = 0.234 s

We need to find the magnitude of the average angular acceleration of the disk. It is given by change in angular velocity per unit time. So,

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So, the magnitude of the average angular acceleration of the disk is 4139.74\ rad/s^2.

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