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tensa zangetsu [6.8K]
3 years ago
7

Explain how a covalent compound can be a strong electrolyte and give an example.

Physics
1 answer:
Luden [163]3 years ago
4 0

Explanation:

Any Substances which dissolute in water yield to ions are called electrolytes. The electrolytes can be both covalent or electrovalent compounds. Polar covalent compounds can be strong electrolyte. Some common examples of covalent electrolytes are HCl, HBr, HI , etc. Solar compounds dissolve in polar solvent ( water).

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Andrew [12]

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B. An electric current into a magnetic field

Explanation:

The generation of electrical power requires relative motion between a magnetic field and a conductor. In a generator, mechanical energy is converted into electrical energy. The electricity produced by most generators is in the form of alternating current.

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Calculate the horizontal force that must be applied to a 1,300 kg vehicle to give it an acceleration of 2.6 m/s² on a level road
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The answer is 3,380 N
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There is friction between the block and the
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Answer:

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3 years ago
A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for
Monica [59]

Question

A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g

Part (b) what is the angle of banking of the highway? Give your answer in degrees

Answer:

a. Equation of Tangent

tan(θ) = v²/rg

b. Angle of the banking highway

θ = 0.087°

Explanation:

Given

Radius of the curve = r = 330m

Acceleration of gravity = g = 9.8m/s²

Velocity = v = 8km/h = 8 * 1000/3600

v = 2.22 m/s

a . Write an equation for the tangent of the highway's angle of banking

The Angle is calculated by

tan(θ) = v²/rg

θ = tan-1(v²/rg)

b.

Part (b) what is the angle of banking of the highway? Give your answer in degrees

θ = tan-1(v²/rg)

Substituting the values of v,g and r

θ = tan-1(2.22²/(330 * 9.8)

θ = tan-1(0.001523933209647)

θ = 0.087314873580116°

θ = 0.087°

3 0
3 years ago
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