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MakcuM [25]
2 years ago
14

Tenemos un Cable de cobre de 1 km de longitud cuya sección es de 2 milímetros al cuadrado y queremos saber la resistencia que se

ñalara el ohmetro , su temperatura es de 20°
Physics
1 answer:
Nady [450]2 years ago
7 0

Answer:

8.5 Ω

Explanation:

La resistencia de un material es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

La fórmula de la resistencia (R) viene dada por:

R = ρL/A

Donde ρ es la resistividad del material, L es la longitud del material y A es el área de la sección transversal del material.

Dado que:

L = 1 km = 1000 m, A = 2 mm² = 2 * 10⁻⁶ m², ρ (cobre) = 1.7 * 10⁻⁸ Ωm

Sustituyendo da:

R = 1,7 * 10⁻⁸ * 1000/2 * 10⁻⁶

R = 8.5 Ω

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What is the formula for wheel and axel
aleksley [76]

Answer: MA= Radius of wheel divided by Radius of Axle

Explanation:

3 0
3 years ago
When a honeybee flies through the air, it develops a charge of +20 pC . Part A How many electrons did it lose in the process of
Yuri [45]

Answer:

1.3 × 10⁸ e⁻

Explanation:

When a honeybee flies through the air, it develops a charge of +20 pC = + 20 × 10⁻¹² C. This is a consequence of losing electrons (negative charges). The charge of 1 mole of electrons is 96468 C (Faraday's constant). The moles of electrons representing 20 pC are:

20 × 10⁻¹² C × (1 mol e⁻/ 96468 C) = 2.1 × 10⁻¹⁶ mol e⁻

1 mole of electrons has 6.02 × 10²³ electrons (Avogadro's number). The electrons is 2.1 × 10⁻¹⁶ moles of electrons are:

2.1 × 10⁻¹⁶ mol e⁻ × (6.02 × 10²³ e⁻/ 1 mol e⁻) = 1.3 × 10⁸ e⁻

7 0
3 years ago
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S
Alja [10]

Answer:

16294 rad/s

Explanation:

Given that

M(ns) = 2M(s), where

M(s) = 1.99*10^30 kg, so that

M(ns) = 3.98*10^30 kg

Again, R(ns) = 10 km

Using the law of gravitation, the force between the Neutron star and the sun is..

F = G.M(ns).M(s) / R²(ns), where

G = 6.67*10^-11, gravitational constant

Again, centripetal force of the neutron star is given as

F = M(ns).v² / R(ns)

Recall that v = wR(ns), so that

F = M(s).w².R(ns)

For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence

F = F

G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)

Making W subject of formula, we have

w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]

w = √[{G.M(ns)} / {R³(ns)}]

w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]

w = √[2.655*10^20 / 1*10^12]

w = √(2.655*10^8)

w = 16294 rad/s

7 0
3 years ago
Which of the following is NOT an argument for showing that the Earth must be round: a. during an eclipse of the Moon, the shadow
Hitman42 [59]

Answer: The correct option is option E (the Sun is seen blocking different constellations in the course of a year.

Explanation:

The earth, which is one of the planets of the solar system that supports life, is shperical in shape. The spherical ( round) shape of the earth is marked by the intervening highlands and oceans on its surface.

Evidence to show that the earth is shperical are:

--> The Lunar eclipse: During an eclipse of the Moon, the shadow of the Earth is always seen to be round.

--> Ships Visibility: When ships travel a large distance away, we see their hulls disappear first and their masts disappear last.

-->Altitude of Polaris (North Star): The height of the North Star changes as we travel to different latitudes. That is ,increases as you move toward the North pole, or decreases as you move toward the equator.

--> Aerial photographs: Photographs of the Earth from space always show a round body.

The statement that doesn't prove that the earth is spherical in shape is (the Sun is seen blocking different constellations in the course of a year). The sun is seen in front of stars blocking different constellation in a year because the earth orbits round the sun in a year and not that it is shperical in shape.

4 0
2 years ago
Usain Bolt accelerates at a rate of 3.7
MissTica

Answer:

Explanation:

I think that you to run more than 12 miles

4 0
3 years ago
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