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3 years ago

A student rapidly rubs the palms of both hands

Physics

1 answer:

Lapatulllka [165]3 years ago

3 0

Answer:

Option a

Explanation:

The action of the rapid rubbing of the hands are a result of chemical reaction and responses that takes place inside our body.

When we rub our hands together, that shows the mechanical motion of the body or we can say that while rubbing hands chemical energy gets converted to mechanical energy (here, Kinetic energy).

While rubbing, due to friction, heat is produced and hence we can say that this mechanical energy has transformed to heat energy.

Thus the conversion here is from chemical to mechanical to heat energy.

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A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predete

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Answer:

3.64×10⁻² cm or 3.64×10⁻⁴ m

Explanation:

Current density = Current/cross sectional area.

δ = I/A................. Equation 1

Where δ = current density, I = current, A = cross sectional area of the wire.

make A the subject of the equation

A = I/δ............. Equation 2

Given: δ = 440 A/cm², I = 0.46 A

Substitute into equation 2

A = 0.46/440

A = 1.045×10⁻³ cm²

But,

A = πd²/4................ Equation 2

Where d = diameter of the wire

make d the subject of the equation

d = √(4A/π)................ Equation 3

Given: A = 1.045×10⁻³ cm², π = 3.14

Substitute into equation 3

d = √(4×1.045×10⁻³/3.14)

d = √(1.33×10⁻³)

d = 0.0364 cm

d = 3.64×10⁻² cm or 3.64×10⁻⁴ m

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3 years ago

When drawing ray diagrams involving thin lenses, how many rays (at a minimum) are needed show the image distance and magnificati

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4 years ago

Four fixed point charges are at the corners of a square with sides of length L. Q1 is positive and at (OL) Q2 is positive and at

Ne4ueva [31]

Answer:

A) See Annex

B) Fq₁₂ = K * Q₁*Q₂ /16 [N] (repulsion force)

C) Fq₃₂ = K * Q₃*Q₂ /16 [N] (repulsion force)

D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)

E) Net force (its components)

Fnx = (2,59/64 )* K*Q² [N] in direction of original Fq₃₂

Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂

Explanation:

For calculation of d (diagonal of the square, we apply Pythagoras Theorem)

d² = L² + L² ⇒ d² = 2*L² ⇒ d = √2*L² ⇒ d= (√2 )*L

d = 4√2 units of length (we will assume meters, to work with MKS system of units)

B) Force of Q₁ exerts on charge Q₂

Fq₁₂ = K * Q₁*Q₂ /(L)² Fq₁₂ = K * Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)

C) Force of Q₃ exerts on charge Q₂

Fq₃₂ = K * Q₃*Q₂ /(L)² Fq₃₂ = K * Q₃*Q₂ /16 (repulsion force in the direction indicated in annex)

D) Force of -Q₄ exerts on charge Q₂

Fq₄₂ = K * Q₄*Q₂ / (d)² Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)

E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)

Let´s take the force that Q₄ exerts on Q₂ and Q₂ = Q ( magnitude) and

Q₄ = -Q

Then the force is:

F₄₂ = K * Q*Q / 32 F₄₂ = K* Q²/32 [N]

We should get its components

F₄₂(x) = [K*Q²/32 ]* √2/2 and so is F₄₂(y) = [K*Q²/32 ]* √2/2

Note that this components have opposite direction than forces Fq₁₂ and

Fq₃₂ respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively

In new conditions

Fq₁₂ = K * Q₁*Q₂ /16 becomes Fq₁₂ = K * Q²/ 16 [N] and

Fq₃₂ = K* Q₃*Q₂ /16 becomes Fq₃₂ = K* Q² /16 [N]

Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively

Then over x-axis we subtract Fq₃₂ - F₄₂(x) = Fnx

and over y-axis, we subtract Fq₁₂ - F₄₂(y) = Fny

And we get:

Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2 ⇒ Fnx = K*Q² [1/16 - √2/64]

Fnx = (2,59/64 )* K*Q²

Fny has the same magnitude then

Fny =(2,59/64 )* K*Q²

The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces

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