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3 years ago

Water molecules are formed by covalent

Physics

1 answer:

frutty [35]3 years ago

3 0

Answer:

When covalent compounds dissolve in water, they break apart into molecules, but not individual atoms. Water is a polar solvent, but covalent compounds are usually nonpolar. This means covalent compounds typically don't dissolve in water, instead, it makes a separate layer on the water's surface.

Explanation:

I don't understand the second part but hope this helps.

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An object thats NEGATIVELY charged has more electrons than protons, An object thats POSITIVELY charged has less electrons than protons, An object that NOT charged has the same number of electrons and protons.

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What happens to the kinetic energy of a snowball as it rolls across the lawn and gains mass.

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If the ground is level, then the snowball can never have
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4 years ago

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Sirius A has a luminosity of 26 LSun and a surface temperature of about 9400 K.What is its radius?

wlad13 [49]

Answer

Given,

Sirius A surface temperature,T = 9400 K

Sirius A luminosity,L = 26 L₀

L₀ is the luminosity of sun.

Radius of sun = 695700000 m

Temperature on sun surface = 5780 K

Luminous intensity is given by:-

$L=4 \pi R^{2} \sigma T^{4}$

Now

$\frac{L}{L_{0}}=\frac{4 \pi R^{2} \sigma T^{4}}{4 \pi R_{0}^{2} \sigma T_{0}^{4}}=26$

$\Rightarrow \frac{R^{2} T^{4}}{R_{0}^{2} T_{0}^{4}}=26$

$\Rightarrow R^{2}=26 \times \frac{R^{2} T_{0}^{4}}{T^{4}}=26 \times \frac{695700000^{2} \times 5780^{4}}{9400^{4}}$

$R=1341246640\ m=1.34 \times 10^{9}$

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3 years ago

A thin uniform rod (mass = 0.53 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plan

gladu [14]

Answer:

(a) L = 0·73 m

(b) 4·39 × $10^{-3}$ J

Explanation:

(a) From the figure, consider the torque about the point where the rod is attached because if we consider another point then there will be hinge forces acting on the rod at the point of attachment

Let L m be the length of the rod and β be the angle between the rod and the vertical

Let α be the angular acceleration of the rod

As the force of gravity acts at the centre so from the figure, the torque about the point of attachment will be 0·53 × g ×(L ÷ 2) ×sinβ

Assuming that the value of amplitude of this oscillation to be small

As torque = moment of inertia × angular acceleration

0·53 × g ×(L ÷ 2) ×sinβ = ((0·53 × L²) ÷ 3) × α (∵ moment of inertia of the rod from the point of attachment)

<h3>For small oscillations, α = ω² × β</h3>

After substituting the value of α and solving we get

ω = √((3 × g) ÷ (2 × L))

Time period = (2 × π) ÷ ω = (2 × π) ÷ √((3 × g) ÷ (2 × L))

∴ (2 × π) ÷ √((3 × g) ÷ (2 × L)) = 1·4

Substituting the value of g as 9·8 m/s² and solving we get

L = 0·73 m

(b) At the maximum amplitude condition the velocity will be 0 and potential energy will be maximum and maximum kinetic energy will be attained at the lowest point and hinge forces will not do work as the point of attachment is not moving

∴ Taking the reference for finding the potential energy as the lowest point

<h3>Maximum potential energy = Maximum kinetic energy </h3><h3>As total energy is constant, since there is no dissipative force</h3>

Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 (∵ increment in height is (L × (1 - cosβ)) ÷ 2

∴ Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 After substituting the value we get

Maximum potential energy = 4·39 × $10^{-3}$ J

∴ Maximum kinetic energy = 4·39 × $10^{-3}$ J

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3 years ago