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Vedmedyk [2.9K]
4 years ago
13

A scientist kept 1,000 grams of a radioactive material in a container. After 66 days, he observed that 125 grams of the radioact

ive material was left in the container. Based on the observation, which of these is the most likely inference drawn by the scientist?
Select one:
a. About 500 gram of the material decayed in 33 days.
b. About 500 gram of the material decayed in 44 days.
c. About 62.5 gram of the material will be left in the container after 88 days.
d. About 62.5 gram of the material will be left in the container after 110 days.
Physics
1 answer:
bixtya [17]4 years ago
8 0
The sample appears to have gone through 3 half-lives
1st half life: 1000 to 500 g
2nd half life: 500 to 250 g
3rd half life: 250 to 125 g
The duration of a half-life, therefore, can be inferred to be 66 ÷ (3) = 22 days. 

After a 4th half life, there will be 125÷2= 62.5 g. 
At this point, an additional 22 days will have passed, for a total of 88 days.
Answer is C. 

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Answer:

d. Direction and magnitude

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The two components of a vector are its magnitude and direction.

Magnitude is the quantity of the substance

Direction is the path.

  • Other quantities are called scalar quantities.
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Examples of vector quantities are velocity, displacement, acceleration.

4 0
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A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second.
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Assuming Earth's gravity, the formula for the flight of the particle is: 

<span>s(t) = -16t^2 + vt + s = -16t^2 + 144t + 160. </span>

<span>This has a maximum when t = -b/(2a) = -144/[2(-16)] = -144/(-32) = 9/2. </span>

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7 0
4 years ago
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What do you think will happen to the potential energy of the ball if the mass of the ball remains the same and placed in a highe
lapo4ka [179]

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Potential energy of the ball would increase

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In your question, the value for height is increased when the object (ball) is placed in a higher position. This creates a net positive change in the right-hand side of the formula and since the value of mass and gravity doesn't change because gravity is a constant (meaning it's always 9.8N/kg) and the mass is said to be kept the same, the ball's potential energy would increase.

4 0
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A 330-ohm resistor is connected to a 5-volt battery. The current through the resistor is
klasskru [66]

Question :-

  • A 330 Ohm Resistor is connected to a 5 Volt Battery . What is the Current through the Resistor ?

Answer :-

  • Current of the Battery is 66 Ampere.

Explanation :-

As per the provided information in the given question, The Resistance is given as 330 Ohm . The Voltage is given as 5 Volt . And, we have been asked to calculate the Current .

For calculating the Current , we will use the Formula :-

\bigstar \:  \:  \:  \boxed{ \sf{ \: Current \:  =  \:  \dfrac{Voltage}{Resistance}  \: }}

Therefore , by Substituting the given values in the above Formula :-

\dag \: \: \: \sf {Current \: = \: \dfrac {Voltage}{Resistance} }

\longmapsto \: \: \: \sf {Current \: = \: \dfrac {5}{330} }

\longmapsto \: \: \: \sf {Current \: = \: \dfrac {1}{66} }

\longmapsto \: \: \: \textbf {\textsf {Current \: = \: 66 }}

Hence :-

  • Current = 66 Ampere .

\underline {\rule {180pt} {4pt}}

Additional Information :-

\Longrightarrow \: \: \: \sf {Voltage \: = \: Current \: \times \: Resistance}

\Longrightarrow \: \: \: \sf {Current \: = \: \dfrac {Voltage}{Resistance} }

\Longrightarrow \: \: \: \sf {Resistance \: = \: \dfrac {Voltage}{Current} }

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