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How much should a 5ft 2in woman weigh?

Updated: 8/11/2023
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Wiki User

12y ago

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There is no set answer for the 'average' weight of a 5' 3" woman. Body weight is a misleading measure. Since people have different builds, different people of the same height and sex can have the same percentage of body fat and yet have different body weights. Therefore, what is important is body composition. The reason for this is that muscle weighs more than fat. What counts more than your body weight is your percentage of body fat. As a female, your percentage of body fat should be about 15 to 20% for good health. To determine your percentage of body fat at home, obtain some plastic body fat calipers and do a simple skin-fold test. Body fat calipers are not expensive and they are readily available to buy. ----

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Wiki User

15y ago
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Wiki User

12y ago

Well, for the weight to be considered "average" it would be 102-139 lbs.

According to an app on my phone, being UNDER 102 (if your 5'2) is considered underweight and being OVER 139 is considered, well, fat.

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Wiki User

12y ago

Male and 5'2 should weight about 135 - 145 pounds.

Female and 5'2 should weight about 118 - 135 pounds.

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Wiki User

15y ago

it depends... i am 5'2" and i weigh about 108lbs

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Wiki User

12y ago

108-121 lbs

48-54 kg

7.7-8.6 st

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Wiki User

13y ago

it depends on your body

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Wiki User

12y ago

Between 52 and 58 kg

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Anonymous

Lvl 1
3y ago

Acvording to BMI calculators a healthy weight ranges from 102 to 134.

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Q: How much should a 5ft 2in woman weigh?
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A machine gun is mounted on the top of tower 100m high at what angle should the gun be inclined to cover a max. range of firing on ground below The muzzle speed of bullet is 150 meters p sec?

Here,u = 150 ms-1And, Taking g = 10 m-2 (approximately)Let 'θ' be the angle at which the machine gun should fire in order to cover a maximum distance.Then , the horizontal component of velocity = 150 cos θAnd, the vertical component of velocity = 150 sin θIf 'T' is the time of flight, then ,Horizontal range, R = (150 cos θ ) x T .The gun is mounted at the top of a tower 100 meters high .Let us regard the positive direction of the position-axis as to be along the line from the top of tower in downward direction.For motion along vertical :Initial Velocity = -150 sin θ;Distance covered = + 100 mAnd, acceleration = + 10 ms-2In time 'T' , the machine gun shot will reach maximum height and then reach the ground.Now,S = ut + 1/2 at2Therefore, +100 = ( -150 sin θ ) T + 1/2 x T2Or, T2 - ( 30 sin θ )T - 20 = 0Therefore, T = - ( - 30 sin θ ) ± { ( - 30 sin θ)2 - 4 x 1 x (-20) }1/2/2= 30 sin θ ± (900 sin2 θ + 80 )1/2/2Or, T = 15 sinθ ± (225 sin2θ + 20)1/2Now, range will be maximum, if time of flight is maximum .Therefore, choosing positive sign ,we haveT = 15 sin θ + ( 225 sin2θ+ 20 )1/2Hence, horizontal range covered,R = 150 cos θ {15 sin θ+ ( 225 sin2θ + 20 )}1/2The horizontal range is maximum, when θ = 45o.But in the present case , the machine gun is mounted at height of 100 m .Therefore, R will not be maximum for θ = 45o.It will be maximum for some value of θ close to 45o.If we calculate values of R by setting θ = 43o, 43.5o, 44o, 45o, 46o and 47o, the values of R come out to be 2347 m, 2347.7 m, 2348 m, 2346 m, 2341 m and 2334 m respectively.Thus R is maximum for value of θ some where between 43.5o and 44o.Therefore , the mean value of θ = (43.5o + 44o)/2 = 43.75o.The gun should be inclined at 43.75o to cover a maximum range of firing on the ground below.