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lara31 [8.8K]
3 years ago
9

If a woman weighs 125 lb, her mass expressed in kilograms is x kg, where x is

Physics
2 answers:
almond37 [142]3 years ago
8 0
Weight of the women in pounds = 125 lb 
1 kg is heavier than 1 pound and 1 Kilogram is same as 2.20462 pounds.  
So 125 lb gives 125 / 2.20462 kilo grams = 56.699 kg and the value is less than 125.
adelina 88 [10]3 years ago
3 0
The first thing you should know to solve this problem is the conversion of pounds to kilograms:
 1lb = 0.45 Kg
 We can solve this problem by a simple rule of three
 1lb ---> 0.45Kg
 125lb ---> x
 Clearing x we have:
 x = ((125) / (1)) * (0.45) = 56.25 Kg.
 Answer
 her mass expressed in kilograms is 56.25 Kg.
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What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?
densk [106]
Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get 
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Tf = -30°c
7 0
3 years ago
Read 2 more answers
In which city rusting is a problem delhi or mumbai
ss7ja [257]

Answer:

Mumbai

Explanation:

Because Mumbai is a coastal city

due to high humidity,vigorous rusting damages all the iron structures

∴ rusting is a major problem in Mumbai

3 0
3 years ago
A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on
n200080 [17]

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

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6 0
3 years ago
Increase the slit width to 1050 nm. What happens to the width of the central bright fringe? (Always remember to wait a few secon
djyliett [7]

Answer:

b.

Explanation:

In case of Single Slit, diffraction will occur.

Then In Single slit Diffraction, width of central fringe is

x_c= 2D\lambda/a


where D= distance b/w screen and slit

a= slit width

\lambda = wavelength

Thus if Screen width increases keeping other factors same then width of central fringe becomes narrower as

x_c\propto 1/a

On increasing the slit width the central bright fringe width The width of the central bright fringe becomes narrower.

3 0
4 years ago
The lons entering the mass spectrometer have the same charges. After being accelerated through a potential difference of 8.20 kV
Ratling [72]

The calculated magnitude is  6.73 x 10³ V/m.

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radius= 19.4/100=0.194 m

total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194

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therefore Magnitude= 8.20 x 10³ / 1.218

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Learn more about Magnitude here-

brainly.com/question/15681399

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4 0
1 year ago
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