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4 years ago

What is different between an ionic bond and a covalent bond?

Physics

2 answers:

Lostsunrise [7]4 years ago

6 0

Ionic bonds will be a metal + a nonmetal and electrons are transferred from the metal to the nonmetal.

A covalent bond will be nonmetals only. nonmetals do not give up their electrons so electrons are shared in this process.

Natali [406]4 years ago

5 0

Ionic Bonding: The formation of an Ionic bond is the result of the transfer of one or more electrons from a metal onto a non-metal. Covalent Bonding: Bonding between non-metals consists of two electrons shared between two atoms.

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A car has two horns, one emitting a frequency of 199 hz and the other emitting a frequency of 203 hz. what beat frequency do the

andriy [413]

Beat frequency, fb = |f2-f1|

That is, beat frequency is the absolute difference between two frequencies. Is is as a results of destructive and constructive inferences.

Therefore, in this case:

fb = 203 - 199 = 4 Hz

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3 years ago

Why would an insurance company want people to avoid risk, like Nick Cannon having to avoid skydiving?

Lina20 [59]

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3 years ago

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What is the time period of a simple harmonic oscillator with a mass of 0.3kg and a force constant of 5N/m?

ikadub [295]

Answer: 6N/m

Explanation:

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3 years ago

A fireman standing on a 15 m high ladder operates a water hose with a round nozzle of diameter 2.02 inch. The lower end of the h

Nataly_w [17]

Answer:

v₁ = 1,606 10⁴ m / s

Explanation:

For this exercise we must use Bernoulli's equation, let's use index 1 for the nozzle on the stairs and index 2 the pump on the street

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² +ρ g y₂

The pressure when the water comes out is the atmospheric pressure

P₁ = P_atm

The difference in height between the street and the nozzle on the stairs is

y₂-y₁ = 15 m

Now let's use the continuity equation

v₁ A₁ = v₂ A₁

The area of a circle is

A = π r² = π (d/2)²

v₁ π d₁²/ 4 = v₂ π d₂²/ 4

v₂ = v₁ d₁² / d₂²

Let's replace

P₂-P_atm + ½ ρ [ (v₁ d₁² / d₂²)²- v₁² ] + ρ g (y2-y1) = 0

P₂- P_Atm + ρ g (y₂-y₁) = ½ ρ v₁² [1- (d₁/d₂)⁴]

v₁² [1- (d₁/d₂)⁴] = (P₂-P_atm) ρ / 2 + g (y₂-y₁) / 2

Let's reduce the magnitudes to the SI system

d₂ = 3.37 in (2.54 10⁻² m / 1 in) = 8.56 10⁻² m

d₁ = 2.02 in = 5.13 10⁻² m

Let's calculate

v₁² [1- (5.13 / 8.56) 4] = 449.538 10³ 10³/2 -9.8 15/2

v₁² [0.8710] = 2.2477 10⁸ - 73.5 = 2.2477 10⁸

v₁ = √ 2.2477 10⁸ /0.8710

v₁ = 1,606 10⁴ m / s

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3 years ago

You are observing a spacecraft moving in a circular orbit of radius 100,000 km around a distant planet. You happen to be located

Natalija [7]

To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:

$v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})$