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swat32
3 years ago
14

Why does the number 54,289.00 have seven significant figures and not five

Physics
2 answers:
nadya68 [22]3 years ago
4 0
The 2 spots after the decimal still counts toward the total  the figures 
Andreyy893 years ago
3 0

thought it was 5 at first too. the 00 past the decimal at significant because it's expressing it is exactly to that point. sort of confusing.
say you measured a pencil with a ruler, and say it was 5" long roughly. you can only scientifically put down that the pencil is 5 inches not 5.0000 because your ruler doesn't go that many decimal places. when something is absolute is the only time you can add .0000000000000000000000 to it and it still be correct. an example of something that is absolute is like you have 10 fingers, 10 is absolute because it's exact 10.0000000000000000=10 fingers it's not any more and not any less.
Those two zeros are included after the decimal so they aqre significant. The rule is all final zeros after the decimal are significant
Since there is a decimal point present, start with the leftmost nonzero didgit. Count that digit and every other digit to the right of it.
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Answer:

solution:

to find the speed of a jogger use the following relation:  

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)  

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619  

=

t

+

C

Here C =constant of integration.   

x

=

0  at  t

=

0

, we get:  C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619  =

t

−

4.7619

.

.

.

.

.

.

.

.

.

(

2

)

Here

x  is measured in miles and  t  in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),    

     to get:

      = −

4.7619  

      =  

1

−

4.7619

      = −

3.7619

  or  x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in   m

i

l

/

differentiate  

     equation (1) with respect to time.

     we have to eliminate x from the equation (1) using equation (2).  

     Eliminating x we get:

     v

=

7.5×

     Now differentiating above equation w.r.t time we get:

      a

=

d

v/

d

t

       =

−

0.675

/

      At  

      t

=

0

      the joggers acceleration is :

       a

=

−

0.675

m

i

l

/

        =

−

4.34

×

f

t

/  

(c)  required time for the jogger to run 6 miles is obtained by setting  

        x

=

6  in equation (2).  We get:

        −

4.7619

(

1

−

(

0.04

×

6  )

)^

7

/

10=

t

−

4.7619

         or

         t

=

0.832

h

r

s

6 0
3 years ago
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