Question

help me pleeeeeeeeeeeeeeeeeease

Answer 1

Note that the gravity acceleration at equatorial sea level is g = 32.088 fps^2 and that its variation is -0.003fps^2 per 1000 ft ascent. Find the height in miles above this point for which a. the gravity acceleration becomes 30.504 fps^2 b. the weight of a given man is decreased by 5 % c. what is the weight of a 180 lb(mass) man stop the 29,131 ft Mt. Everest in Tibet, relative to the point

Answer 2

in thermodynamics/physics

Answer 3

@baru

Answer 4

for section a:

what is the change in g betwen the 2 values 32.088 & 30.504 ?

Answer 5

-1.284 f/s^2

Answer 6

I assume that is a typo - it is -1.584ft/s^2

now g decreases by -.003 for every 1000 ft eleveation

so divide your figure by .003 to find how many 1000's ft you need to go to reduce by -1.584

Answer 7

yes it is a typo

Answer 8

what formula you used to get the h?

Answer 9

h will be 528,000 ft

Answer 10

ok - good - now convert that to miles

Answer 11

wait whats the formula

Answer 12

look it up

Answer 13

no i mean for example
h= a/k like that

Answer 14

how did you get the answer without the equation?

so divide your figure by .003 to find how many 1000's ft you need to go to reduce by -1.584

Answer 15

the formula like F = ma..

Answer 16

a mi = 5280 ft
so 9528,000 ft)(1mi/5280) = 100 miles

Answer 17

528,000ft(1mi/5280 ft = 100 miles

Answer 18

OK - that is the answer to section a


now weight = ma
but the mass of the man soenst change
so to reduce his weight by 5% you must reduce a by 5%

so what is the value for a that is 5% less than 32.088 fs^-2?

Answer 19

theres no given mass

Answer 20

but the mass of the man doenst change
so to reduce his weight by 5% you must reduce a by 5%

Answer 21

i cant understand

Answer 22

it asks you to find the height at which his weight is reduced by 5%

since the mass doesnt change you must find the height at which the acceleration has changes by -5%

so what is the VALUE of a if it is 5% less than 32.008?

Answer 23

32.038?

Answer 24

if for example you reduce 100 by 5% what do you get?

Answer 25

i get 5

Answer 26

no - you get 95

so what do you get if you reduce 32.008 by 5%?

Answer 27

i get 27. 088

Answer 28

no - you have reduce it by 5 - not by 5%

if you cannot work out percentages then you don't have the background sklills to solve this equation.

Go and revise % calculations and find what 32.008 -5% is

Answer 29

i put it in calcu 32.088 - 5% (which is 0.5) the answer is 31.588

Answer 30

0.05 i means

Answer 31

yeah - like I said - you don't seem to have the arithmentic knowledge to work this out.

Instead of putting it in oyur calculator - look up how to work it out, tehn com back when you have the value of 32.008 when reduced by 5%

Answer 32

It isn't reasonable to ask for help on a question that requires the helper to go back to basic arithmentic or fundamental requirements.

You need to build on your background, and I have to assume that before you learnt this type of question you learnt how to work out %

Answer 33

let me rep[hrase the question

what is 5% of 32.008?

Answer 34

1.6044

Answer 35

so if you reduce the accleration by 5% what is the new value?

Answer 36

30.4836

Answer 37

OK good

so now section b is exactly the same as section a (but with a differnt final acceleration)

so use the same method to find th eheight where the accelertion is 30.4836

Answer 38

\[= \frac{ 30.4836 }{ -0.003 } diide 1000\]

Answer 39

no - that is not what oyu did for section a

just slow down a bit and be methodical

Answer 40

its now -10161200 ft

Answer 41

refer to my very first post, and do the same for section b as follows:

for section a:
what is the change in g betwen the 2 values 32.088 & 30.504 ?

Answer 42

-1.584 fts2

Answer 43

and in b is -30.4836

Answer 44

wait it sould be 30.4836 - 32.088

Answer 45

OK - now complete the same as you did for a

Answer 46

534800

Answer 47

in miles?

Answer 48

101. 3 mile

Answer 49

hmm - not what I get

I think you need to look again at the 5% value

Answer 50

but 1 miles is equal to 5280 ft

Answer 51

yes - but you went wrong earlier - I diudn't check your numbers

do 5% of 32.008 (you wrote it wron above)

then subtract that to get teh new value for a

(you have the method correct now - but th enumbers are wrong - keep going.. :-)

Answer 52

in the question it is 32.088

Answer 53

not the 32.008

Answer 54

sorry - my mistake

I think maybe your answer is correct- wait a minute..

Answer 55

okay i wait

Answer 56

32.088*5% = 1.6044

1.6044/.003 = 53480 ft = 101.29 miles

Answer 57

yes, i round off it

Answer 58

OK - good

Answer 59

so what about the c

Answer 60

so what is acceleration at 29,131 ft?

Answer 61

a=161.84

Answer 62

no - how many '000s of feet is it?

Answer 63

Hey. Whats the question you need help on

Answer 64

@warfare432 c. what is the weight of a 180 lb(mass) man stop the 29,131 ft Mt. Everest in Tibet, relative to the point

Answer 65

the original question is Note that the gravity acceleration at equatorial sea level is g = 32.088 fps^2 and that its variation is -0.003fps^2 per 1000 ft ascent. Find the height in miles above this point for which a. the gravity acceleration becomes 30.504 fps^2 b. the weight of a given man is decreased by 5 % c. what is the weight of a 180 lb(mass) man stop the 29,131 ft Mt. Everest in Tibet, relative to the point

Answer 66

the question to be answered is letter c @Directrix

Answer 67

@timo86m

Answer 68

Sorry, I don't know this.

Answer 69

@Roxy.girl help me find the acceleration in letter c

Answer 70

please post one question at a time. :)

Answer 71

:O so much work!

Answer 72

I M SO CONFUSED BY DIS STUFF DX

Answer 73

First find the variation that corresponds to that height relative to sea level
29131ft / 1000 = 29.131 ft, so the variation in g relative to sea level is
-0.003fps^2 x 29.131 = -0.08739. Thus g at that height is sea level g minus with the variation
32.088-0.08739 = 32.00061
W = mg (now I dunno how American units work out in relation to pounds and feet) but we’ll just use a relative measure. The difference in g by ratio is now 32.00061/32.088 of original weight, since mass is maintained which is 99.73% to 4 significant figures.
So 99.73% of 180 lbs = 179.5 lbs. So he loses half a pound in weight (not in mass)