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IceJOKER [234]
3 years ago
13

A.Substitute in the units for each one and combine like terms.

Physics
1 answer:
storchak [24]3 years ago
8 0
IDK     ghjfnhgfjmrmhjgfhgfmmfh
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A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20
Klio2033 [76]

Answer:

\boxed{{\boxed{\blue{ 12.5}}}}

Explanation:

Given, for girl : Weight or force;

\rm \: F_1 = 50 \: kgf

Area of both heels;

\rm \: A_1 =  \; 2 ×1 \;  cm^2 = 2  \: cm^2

\rm \: Pressure \:  P_1  =  \cfrac{F_1}{ A_1 }  =  \dfrac{50 \: kgf}{2 \: cm {}^{2} }  = 25 \: kgf \: cm {}^{ - 1}

For elephant, Weight = Force \rm F_2 = 2000 kg•f

Area of 4 feet;

\rm \: A_2  = \; 4 \times 250 \;  cm^2 = 1000 \:  cm^2

\rm \: Pressure \:  P_2 = {F_2}/{A_2} \;  = \cfrac{2 \cancel{0 00 }\:  kgf}{1 \cancel{000} \: cm^2} =  2 \: kgf \: { \:cm}^{- 1}

Now;

\rm  = \dfrac{Pressure \:  Exerted  \: by  \: the \:  Girl}{Pressure  \: exerted  \: by \:  the  \: elephant}

=  \rm \: P_1/P_2

\implies    \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } =  \rm\cfrac{25 \:  \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

3 0
2 years ago
Distinguish between directly proportional and inversely proportional with an example of each
RideAnS [48]

Consider two variables said to be "inversely proportional" to each other. If all other variables are held constant, the magnitude or absolute value of one inversely proportional variable decreases if the other variable increases, while their product (the constant of proportionality k) is always the same.

4 0
3 years ago
A large piece of jewelry has a mass of 130.8 g. A graduated cylinder initially contains 47.7 mL water. When the jewelry is subme
Shkiper50 [21]

Answer: The density of this piece of jewelry is 8.90g/cm^3

Explanation:

To calculate the density, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Mass of piece of jewellery = 130.8 g

Density of piece of jewellery = ?

Volume of piece of jewellery =( 62.4-47.7 ) ml = 14.7 ml = 14.7cm^3   1cm^3=1ml

Putting values in above equation, we get:

\text{Density of piece of jewellery}=\frac{130.8g}{14.7cm^3}=8.90g/cm^3

Thus density of this piece of jewelry is 8.90g/cm^3

8 0
3 years ago
If a 5 N force pushes a 20 kg mass on a frictionless surface, how fast is the mass going in 5 seconds.
stira [4]

Explanation:

Let me know if you have questions

3 0
3 years ago
What is the formular for lmpulse​
Brums [2.3K]

Answer:

the change in momentum = Force x change in time

3 0
2 years ago
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