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icang [17]
3 years ago
7

Assignment S of write the Symbol Told, mercury and Cooper, Iron, Lead

Physics
1 answer:
m_a_m_a [10]3 years ago
3 0

Answer:

Check explanation

Explanation:

Gold - Au (Aurum)

Mercury - Hg (Hydrargyrum)

Copper - Cu (Cuprum)

Iron - Fe (Ferrum)

Lead - Pb (Plumbum)

These elements in the periodic table are some of the elements represented by letters not in line with their names.

This is because, these elements were known in ancient times and therefore, they are represented by letters from their ancient names.

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A point charge is placed 100 m from a 6 uC charge generating an electric field. What is the
kvv77 [185]

The strength of the electric field is 5 N/C

Explanation:

The magnitude of the electric field produced by a single-point charge is given by:

E=\frac{kQ}{r^2}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

Q=6 \mu C = 6\cdot 10^{-6}C is the charge producing the field

r = 100 m is the distance from the charge at  which we want to calculate the field

Substituting into the equation, we find the s trength of the electric field:

E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C

Learn more about electric field:

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3 years ago
A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. A
Ghella [55]

The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

That is,

• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point

• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium

so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

<em>W</em> = ∆<em>K</em> = <em>K</em>

where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So

<em>W</em> = 1/2 <em>mv</em> ²

where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get

<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s

8 0
3 years ago
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Butoxors [25]

Explanation:

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2 years ago
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Consider a perfectly reflecting mirror oriented so that solar radiation of intensity i is incident upon, and perpendicular to, t
podryga [215]
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1. Different amount of energy is transferred to different parts of an electric circuit. *
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Answer:

True

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