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maryam

@aishu_fan

Voyager

Posted: 16 years ago

#101

3 questions i need help in,i dotn get at all,my brain hurts from reading the lesson!

#1) earth has a mass of 5.97x10^24kg and a radius of 6.38 x105m while saturn has a mass of 5.68x1026kg and a radius of 6.03x107. find the weight of a 65.0 kg person at the followin locations:

A.ont he surface of earth

B.1000 km above the earth surface

C.on the surface of saturn

D.1000km above the surface of saturn

ok so A and C are basically the same so i think if u explain a ill get c,but i dont get the other 2?

ok for A im not sure but this is what i think is should do:

formula:fg=G m1m2(or it can be written mm)/r^2 or we could use other forumals i dont no

but for tht one rent u supposed to find the mass?but then wat about fg u no? im confused?i think im looking at it all wrong...ugh!helllp!

#2what is the magnitude of g at a height above earth's surface where free fall acceleration equals 6.4 m/s^2?

#3 suppose the value of G has just been discovered. use the value of G and an approximate value for the earth's radius(6.38x10^6m)to find an apprximitation for earth's mass?

I DOTN GET It...AT ALL?!? HEEEEEEEELLLP PLZZ

THNK U

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akhl

@akhl

Dazzler

Posted: 16 years ago

#102

1) Mass of Earth Me = 5.97x10^24 kg

Radius of Earth Re = 6.38 x10^5 m

Mass of Saturn Ms = 5.68x10^26 kg

Radius of saturn Rs = 6.03x10^7 m

Mass of person M = 65.0 kg

A)Weight = G*M*Me/Re^2

Can you substitute values and get the answer?

B)Altitude h = 1000 km = 10^6 m

Weight = G*M*Me/(Re + h)^2

Another method

Weight = weight on surface * Re^2/(Re + h)^2

You have calculated weight on surface in (A).

C) Weight = G*M*Ms/Rs^2

D) Altitude h = 1000 km = 10^6 m

Weight = G*M*Ms/(Rs + h)^2

Another method: -

Weight = weight on surface * Rs^2/(Rs + h)^2

You have calculated weight on surface in (C).

2) g = 6.4 m/s^2

(Tip: Free fall acceleration means g.)

3) g = 9.8 m/s^2

G = 6.67 * 10^-11 N-m^2/kg^2

R = 6.38 * 10^6 m

M = ?

g = GM/R^2

Or M = g * R^2/G

= 9.8 * (6.38 * 10^6)^2/(6.67 * 10^-11)

= 5.98 * 10^24 kg

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Anu

@sweet_si_anu

Voyager

Posted: 16 years ago

#103

Help me with my fashion design assignment plz!!

Hi! I have been given an assignment where I've to select one texture from any magazine or anywhere and recreate it using different materials like pasta, pencil shavings, wool, thread, paper cuttings , straws, matchsticks, etc .... and I'm confused as in what to select as a texture,,, /I've thought of using the following pic(marble jaali)...but I'm not very sure whether it can be used as texture or not..please suggest... Thanks in advance!!

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maryam

@aishu_fan

Voyager

Posted: 16 years ago

#104

Water is to be pumped to the top of the Empire State Building, which is 1200 ft high. What gauge pressure is needed in the water line at the base of the building to raise the water to this height?

i dont get it? wat forumla do i use? and how?

ok so in our lesson we r given these forumals: p=P0+pgh

absloute pressure= atmospheric pressure+(density Xfree fall accel.Xdepth)

P=F/a

Pinc=f1/a1=f2/a2

so can u liek expalin it ?thnks!

Edited by aishu_fan - 16 years ago

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akhl

@akhl

Dazzler

Posted: 16 years ago

#105

By definition, gauge pressure = absolute pressure - atmospheric pressure

Let atmospheric pressure = Po,

h = height of the building,

p = density of water

Then pressure at the base of the building = Po + pgh

Gauge pressure = Po + pgh - Po = pgh

p = 1000 kg/m^3

h = 1200 ft

1 ft = 12 inch and 1 inch = 2.54 cm

Therefore h = 1200 * 12 * 2.54 cm = 12 * 12 * 2.54 m = 365.76 m

Gauge pressure = 1000 * 9.8 * 365.76 = 3584448 Pa

This is approx. 3.6 MPa

You can convert into another unit if needed.

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maryam

@aishu_fan

Voyager

Posted: 16 years ago

#106

when a submarine dives to a depth of 5.0 X10^2m how much pressure,in pa,must its hull be able to withstand?

how many times larger is this pressure than the prssure at the surface? plz help on this one too

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maryam

@aishu_fan

Voyager

Posted: 16 years ago

#107

o yea and aap mahan ho! thnk very very much for helping me!

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akhl

@akhl

Dazzler

Posted: 16 years ago

#108

Problem: when a submarine dives to a depth of 5.0 X10^2m how much pressure,in pa,must its hull be able to withstand? how many times larger is this pressure than the prssure at the surface? plz help on this one too

Solution: Pressure at the surface = atmospheric pressure Po = 10^5 Pa

Height h = 5.0 * 10^2 m

Density of water p = 10^3 kg/m^3

Let pressure at submarine = P

P = Po + pgh = 10^5 + 10^3 * 9.8 * 5.0 * 10^2

= 10^5 + 49 * 10^5

= 50 * 10^5

= 5.0 * 10^6 Pa

P/Po = 5.0 * 10^6/10^5 = 50

Ans: The hull of the submarine has to withstand a pressure of 5.0 * 10^6 Pa

This is 50 times the pressure at the surface.

Note: Some people use Po = 101325 Pa and some use simplified value of 100000 i.e. 10^5 Pa. I have used 10^5 Pa. Use the one that your teacher has said.

Edited by akhl - 16 years ago