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2 years ago

A manufacturer supplies plastic cups that are placed under the legs of the chair

1 answer:

6 0

A manufacturer supplies plastic cups that are placed under the legs of the chair. The manufacturer's claim is that there will be no marks of chair due to pushing or pulling on the floor.

<h3>Who is a manufacturer?</h3>

A manufacturer is a person who converts his idea of developing and constructing any item using some set of machines.

A manufacturer supplies plastic cups that are placed under the legs of the chair.

When there is a relative motion between two objects, friction force causes the opposition to this motion. Between chair and floor, there will be high magnitude of force. But if we insert plastic cups in the legs of chair, then the friction between cup and floor will be less. This keeps the floor protected form any scratches.

Learn more about manufacturer.

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What is/are the products in the chemical equation?

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2 years ago

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Newer telephone circuits, built during the last decade, offer higher quality because they were built using analog transmission.

hammer [34]

Answer:

Well, newer telephone circuits built during the last decade are based on the digital transmission, not on the analog transmission. So it's the digital transmission circuit that has made the higher quality. Digital circuits converts the voice signals into the binary codes which is then translated again into the voice signal at the receiving end.

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3 years ago

A car with mass m traveling at speed v has kinetic energy k. what is the kinetic energy of a second car that has the same mass m

vampirchik [111]

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3 years ago

When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl

zheka24 [161]

Answer:

I = 21.13 mA ≈ 21 mA

Explanation:

I₁ = 5 mA

L₁ = L₂ = L

V₁ = V₂ = V

ρ₁ = 1.68*10⁻⁸ Ohm-m

ρ₂ = 1.59*10⁻⁸ Ohm-m

D₁ = D

D₂ = 2D

S₁ = 0.25*π*D²

S₂ = 0.25*π*(2*D)² = π*D²

If we apply the equation

R = ρ*L / S

where (using Ohm's Law):

R = V / I

we have

V / I = ρ*L / S

If V and L are the same

V / L = ρ*I / S

then

(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂

S₁ = 0.25*π*D² and

S₂ = 0.25*π*(2*D)² = π*D²

we have

ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)

⇒ I₂ = 4*ρ₁*I₁ / ρ₂

⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m

⇒ I₂ = 21.13 mA

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