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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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Circular coil 1 has N turns and circular coil 2 has 6N. Coil 1 has area A and coil 2 has area A/4. If Coil 1 has current i runni
solmaris [256]

Answer:

\frac{L_1}{L_2} =\frac{1}{3}

Explanation:

Recall that the formula for an inductance (L) for coil on N turns, are A and current I is given by:

L=\frac{\mu_0\,N^2\,A}{I}

Then, for the first coil we have;

L_1=\frac{\mu_0\,N^2\,A}{I}

and for coil 2 we have:

L_2=\frac{\mu_0\,(6\,N)^2\,(A/4)}{3\,I}

then, the quotient L1/L2 can be written as:

\frac{L_1}{L_2} =\frac{\mu_0\,N^2\,A\,3\,I}{\mu_0\,(6\,N)^2\,(A/4)\,I}=\frac{12}{36} =\frac{1}{3}

7 0
3 years ago
Physical sciences grade 11 prescribed experiment 1
Digiron [165]

Answer:  

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

Explanation:

6 0
3 years ago
Two very small spheres are initially neutral and separated by a distance of 0.30 m. Suppose that 2.50 1013 electrons are removed
quester [9]

Answer:

F = - 1,598 10⁻³ N

Explanation:

Electic strength is given by Coulomb's law

          F = k q₁ q₂  / r²

Where k is the Coulomb constant that is worth 8.99 10⁸ N m²/C², q₁ and q₂ are the charges and r is the distance that separates the electric charges

In this case the charge of the two spheres is the same and of a different sign since when you remove the charge of a sphere that was initially neutral, it is left with that charge removed but of the opposite sign

       q₁ = q₂ = 2.50 10¹³ electrons = 2.50 10¹³ 1.6 10⁻¹⁹

       q₀ = 4.0 10⁻⁶ C

Let's calculate

       F = - 8.99 10⁸ (4.0 10⁻⁶)² / 0.30²

       F = - 1,598 10⁻³ N

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