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RoseWind [281]
3 years ago
9

One shoe has a pointed heel that has a small cross sectional area, and another shoe has a broad heel with a large cross sectiona

l area. Both shoes have the same mass and are made of the same material. Which shoe can cause more pain if the same person wearing it accidently uses their heel to step on another person's foot? And why?
Physics
1 answer:
katovenus [111]3 years ago
5 0

Answer:

Answer:

the shoe which having heel of smaller crossection area.

Explanation:

Pressure is defined as the thrust acting per unit area.

Thrust is the force which acts on the surface perpendicularly.

Pressure = force / Area

Its SI unit is Pascal or Newton per square metre.

As both the shoes have same mass, so force is same but the area of crossection is different for both the shoes.

Teh shoe having heel of small crossection area exerts more pressure on another person's feet then the shoe having heel which has more crossection area.

Explanation:

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borishaifa [10]
<h2>Dimension for cheap enclose = 32.45 ft x 23.52 ft</h2>

Explanation:

Area of rectangular field, A = 830 ft²

Length = l

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So we have

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                    l=\frac{830}{w}

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            Cost for fencing, C = 2 x 2 x w + 3 x 2 x l = 4 w + 6 l

             C=4w+6\times \frac{830}{w}

For minimum cost we have derivative is zero

           dC=4\times dw-6\times \frac{830}{w^2}\times dw\\\\0=4\times dw-6\times \frac{830}{w^2}\times dw\\\\w^2=1245\\\\w=32.45ft\\\\lw=830\\\\l\times 32.45=830\\\\l=23.52ft

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Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to
sergij07 [2.7K]

Answer:

KE=1.2036\times 10^{-12}\ J

Explanation:

Given:

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  • mass of the alpha particle, m=6.64\times 10^{-27}\ kg
  • strength of a uniform magnetic field, B=0.5\ T
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<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

q.v.B=m.\frac{v^2}{r}

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v=\frac{q.B.r}{m}

v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}

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m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m

m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }

m'=6.6533\times10^{-27}\ kg

now kinetic energy:

KE=m'.c-m.c

KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2

KE=1.2036\times 10^{-12}\ J

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