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Lostsunrise [7]
3 years ago
10

7. A cyclist starting from rest accelerates at a rate of 6 m/s until she reaches 30 m/s. How long

Chemistry
1 answer:
Hunter-Best [27]3 years ago
4 0

Answer:

Explanation:

The given values are:

a= 6m/s²

u= 0 m/s

v = 30 m/s

t= ?

The formula is :

a=\frac{v-u}{t}

thus,

t=\frac{v-u}{a}\\t=\frac{30-0}{6}\\t= 5 seconds

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Can anyone help me on this please?
Bad White [126]
This question is testing to see how well you understand the "half-life" of radioactive elements, and how well you can manipulate and dance around them.  This is not an easy question.

The idea is that the "half-life" is a certain amount of time.  It's the time it takes for 'half' of the atoms in any sample of that particular unstable element to 'decay' ... their nuclei die, fall apart, and turn into nuclei of other elements.

Look over the table.  There are 4,500 atoms of this radioactive substance when the time is 12,000 seconds, and there are 2,250 atoms of it left when the time is ' y ' seconds.  Gosh ... 2,250 is exactly half of 4,500 !  So the length of time from 12,000 seconds until ' y ' is the half life of this substance !  But how can we find the length of the half-life ? ? ?

Maybe we can figure it out from other information in the table !

Here's what I found:

Do you see the time when there were 3,600 atoms of it ? 
That's 20,000 seconds.

... After one half-life, there were 1,800 atoms left.
... After another half-life, there were 900 atoms left.
... After another half-life, there were 450 atoms left. 

==>  450 is in the table !  That's at 95,000 seconds.

So the length of time from 20,000 seconds until 95,000 seconds
is three half-lifes.

The length of time is (95,000 - 20,000) = 75,000 sec

                                     3 half lifes = 75,000 sec

Divide each side by 3 :   1 half life = 25,000 seconds

There it is !  THAT's the number we need.  We can answer the question now.

==> 2,250 atoms is half of 4,500 atoms.

==> ' y ' is one half-life later than 12,000 seconds

==> ' y ' = 12,000 + 25,000

         y   = 37,000 seconds  .

Check: 
Look how nicely 37,000sec fits in between 20,000 and 60,000 in the table.

As I said earlier, this is not the simplest half-life problem I've seen.
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7 0
3 years ago
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3 0
3 years ago
A sample of a hydrocarbon is found to contain 7.48g carbon and 2.52g hydrogen. What is the empirical
Aliun [14]

Answer:

CH₄

Explanation:

To determine the empirical formula of the hydrocarbon, we need to follow a series of steps.

Step 1: Determine the mass of the compound

The mass of the compound is equal to the sum of the masses of the elements that form it.

m(CxHy) = mC + mH = 7.48 g + 2.52 g = 10.00 g

Step 2: Calculate the percent by mass of each element

%C = mC / mCxHy × 100% = 7.48 g / 10.00 g × 100% = 74.8%

%H = mH / mCxHy × 100% = 2.52 g / 10.00 g × 100% = 25.2%

Step 3: Divide each percentage by the atomic mass of the element

C: 74.8/12.01 = 6.23

H: 25.2/1.01 = 24.95

Step 4: Divide both numbers by the smallest one, i.e. 6.23

C: 6.23/6.23 = 1

H: 24.95/6.23 ≈ 4

The empirical formula of the hydrocarbon is CH₄.

6 0
3 years ago
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