Help Me With These Simple Formulae Questions?

Hi all, first of all I'd like to thank you for your time. Please answer all three of these questions below in as much detail as you can also including detailed steps. Thank you!

Question 7 & 8
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Question 8 (Review Set)
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3 Answers

Answer:

Question "7" - see below:

Explanation:

For the question "7":

#S=1/2 at^2#

solve for #t#:

#2S=at^2#

#t^2=(2S)/a#

#t=sqrt((2S)/a)#

~~~~~

When #a=8, S=40#

#t=sqrt((2(40))/8)=sqrt(80/8)=sqrt10 sec#

When #a=8, S=10#

#t=sqrt((2(10))/8)=sqrt(20/8)=sqrt(5/2) sec#

Answer:

Question "8" - see below:

Explanation:

Question "8"

#m=m_o/sqrt(1-(v/c)^2)#

Solve for #v#:

#msqrt(1-(v/c)^2)=m_o#

#sqrt(1-(v/c)^2)=m_o/m#

#1-(v/c)^2=(m_o/m)^2#

#(v/c)^2=1-(m_o/m)^2#

#v/c=sqrt(1-(m_o/m)^2)#

#v=csqrt(1-(m_o/m)^2)#

~~~~~

Find #v# as a fraction of #c# such that #m=3m_o#

#v/c=sqrt(1-(m_o/(3m_o))^2)#

#v/c=sqrt(1-(1/3)^2)#

#v/c=sqrt(1-(1/9))#

#v/c=sqrt(8/9)#

#v=csqrt8/3~~.9428c#

~~~~~

Find #v#, with #m=30m_o# and #c~~3xx10^8#

#v=(3xx10^8)sqrt(1-(m_o/(30m_o))^2)#

#v=(3xx10^8)sqrt(1-(1/30)^2)#

#v=(3xx10^8)sqrt(1-(1/900))#

#v=(3xx10^8)sqrt(899/900)~~299,833,287ms^-1#

Answer:

Review Set 8 - see below:

Explanation:

Review Set 8:

With the set of consecutive odd integers, we are given:

#1,3,5,7,...#

The next 3 terms are:

#9,11,13#

The #nth# term can be found:

#2n-1# where #n# is a natural number (#n>0#)

For the pattern of #T_1, T_2, T_3,...#

where #T_1=1/(1xx3), T_2=1/(3xx5), T_3=1/(5xx7)#

following the pattern:

#T_4=1/(7xx9), T_5=1/(9xx11), T_6=1/(11xx13)#

notice that, in the denominator, the smaller of the two numbers is the #nth# term. For instance, for #T_6=1/(11xx13)#, the subscript 6 is #n#, and so the #nth# term is #2n-1=2(6)-1=11#.

Hence, for the 20th term, we'll have the lesser term in the denominator as:

#2n-1=2(20)-1=39#

and therefore:

#T_20=1/(39xx41)#

and

#T_n=1/((2n-1)(2n+1))#