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3 years ago

I NEED HELP ASAP

Physics

2 answers:

LiRa [457]3 years ago

4 0

The object is called a meteor because it is producing Streak of light and has not yet struck earth.

<h3><u>Explanation:</u></h3>

A meteoroid is a celestial object which is very smaller than an asteroid. These objects are produced as a collision impact from mars or moon and float freely in space without any specific orbit. When they come inside the Earth's gravitational field, they are attracted by the Earth's gravity to Earth's crust. These objects in Earth's atmosphere are called meteors. As they travel through Earth's atmosphere, they do face a huge friction from Earth's atmosphere which let them burn and that is visible as the tail of the meteor.

Most of them are so small that they are burnt away in the atmosphere. But some are bigger and they reach the Earth's surface and are called as meteorites.

Maksim231197 [3]3 years ago

3 0

Answer:

Meteor, with a streak of light and it has not yet reached earths atmosphere

Explanation:

This was one of the questions in my mastery test and i got a 100

A meteoroid is a celestial object which is very smaller than an asteroid. These objects are produced as a collision impact from mars or moon and float freely in space without any specific orbit. When they come inside the Earth's gravitational field, they are attracted by the Earth's gravity to Earth's crust. These objects in Earth's atmosphere are called meteors. As they travel through Earth's atmosphere, they do face a huge friction from Earth's atmosphere which let them burn and that is visible as the tail of the meteor.

Most of them are so small that they are burnt away in the atmosphere. But some are bigger and they reach the Earth's surface and are called as meteorites.

(not mine)

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The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

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Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.