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WARRIOR [948]
3 years ago
7

As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o

n the floor by the heel if it has an area of 1.64 cm2 and the woman's mass is 67.0 kg. Express the pressure in pascals. (In the early days of commercial flight, women were not allowed to wear high-heeled shoes because aircraft floors were too thin to withstand such large pressures.)
Physics
1 answer:
Inessa05 [86]3 years ago
7 0

Answer:4.003 MPa

Explanation:

Given

Area of heel A=1.64 cm^2\approx 1.64\times 10^{-4} m^2

Mass of woman m= 67 kg

Pressure is defined as Force applied in over an area such that it is perpendicular to it i.e.

Pressure=\frac{Force}{Area}

P=\frac{F}{A}

P=\frac{mg}{A}

P=\frac{67\times 9.8}{1.64\times 10^{-4}}

P=400.365\times 10^{4} Pa

P=4.003 MPa

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A 4,000 kg truck is moving at +10 m/s hits a 1500 kg parked car which moves off at +10 m/s. What is the velocity of the truck
Igoryamba

Answer:

10m/s

Explanation:

Using the law of conservation of momentums

M1u1+m2u2 = (m1+m2)v

Substitute.

4000(10)+1500(10) = (4000+1500)v

40,000+15,000 = 5,500v

55000 = 5500v

v = 55000/5500

v= 10m/s

Hence the velocity of the truck after Collision is 10m/s

6 0
3 years ago
A 0.49 m copper rod with a mass of 0.15 kg carries a current of 13 A in the positive y direction. Let upward be the positive dir
Helen [10]

Answer:

Magnetic field, B = 0.23 T          

Explanation:

Given that,

Length of the copper rod, L = 0.49 m

Mass of the copper rod, m = 0.15 kg

Current in rod, I = 13 A (in +ve y direction)

When the rod is placed in magnetic field, the magnetic force is balanced by its weight such that :

BIL=mg\\\\B=\dfrac{mg}{IL}\\\\B=\dfrac{0.15\times 9.8}{13\times 0.49}\\\\B=0.23\ T

So, the magnitude of the minimum magnetic field needed to levitate the rod is 0.23 T.

6 0
3 years ago
The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the
MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations.  The half-life formula is P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}} where P_o is the initial amount, k is the length of the half-life, t is the amount of time that has elapsed since the initial measurement was taken, and P is the amount that remains at time t.

P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, y=a b^{t}, where t is still the amount of time, and y is the amount remaining at time t.  The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

y=a b^{t}

(14.2)=ab^{(0)}

14.2=a *1

14.2=a

So, a=14.2, which represents out initial amount of the substance, and our equation becomes: y=14.2 b^{t}

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}

\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}

0.5=b^{8.0252}

\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}

\sqrt[8.0252]{\frac{1}{2}}=b

Recalling exponent properties, one could find that  \left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b, which will give the equation identical to the half-life formula.  However, recalling this trivia about exponent properties is not necessary to solve this problem.  One can just evaluate the radical in a calculator:

b=0.9172535661...

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

y=14.2* (0.9172535661)^t  or y=14.2*(0.5)^{\frac{t}{8.0252}} where y represents the amount remaining at time t.

<u>Solving for the amount remaining</u>

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

y=14.2* (0.9172535661)^{(31.8)}          y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}

y=14.2* 0.0641450581                    y=14.2*(0.5)^{3.962518068}

y=0.9108598257                              y=14.2* 0.0641450581

                                                        y=0.9108598257

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

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If you blow up a balloon, tie it off, and release it, it will fall to the ground.
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Answer:

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