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3 years ago

Elements with positive valences usually ______ electrons

Physics

1 answer:

Leno4ka [110]3 years ago

7 0

The answer is donate, therefore elements with positive valences usually donate electrons

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Describe two of Tesla's inventions that influenced<br> technology and are still in use today.

victus00 [196]

The remote control, and wireless transmissions

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3 years ago

Plzz help mee <br>What do you notice about how we write out the symbols for the different elements?

k0ka [10]

Many of the symbols are the first one or two letters of the element’s name in English. The first letter of a symbol is always written as a capital letter (uppercase) and the second letter as a small letter (lowercase).

For example

(i) hydrogen, H

(ii) aluminium, Al and not AL

(iii) cobalt, Co and not CO.

Symbols of some elements are formed from the first letter of the name and a letter, appearing later in the name.

Examples are: (i)chlorine, Cl, (ii) zinc, Zn etc.

Other symbols have been taken from the

names of elements in Latin, German or Greek.

For example, the symbol of iron is Fe from its Latin name ferrum, sodium is Na from natrium, potassium is K from kalium. Therefore, each element has a name and a unique chemical symbol.

6 0

3 years ago

The electric potential inside a charged spherical conductor of radius R is is given by V = keQ/R, and the potential outside is g

Bumek [7]

Answer:

For outer points of shell

$E = \frac{k_eQ}{r^2}$

Now for inner point of shell

$E = 0$

Explanation:

As we know that out side the shell electric potential is given as

$V = \frac{K_e Q}{r}$

inside the shell the electric potential is given as

$V = \frac{K_e Q}{R}$

now we know the relation between electric potential and electric field as

$E = - \frac{dV}{dr}$

so we can say for outer points of the shell

$E = -\frac{dV}{dr}$

$E = - \frac{d}{dr}(\frac{K_eQ}{r})$

$E = \frac{k_eQ}{r^2}$

Now for inner point again we can use the same

$E = - \frac{dV}{dr}$

$E = - \frac{d}{dr}(\frac{K_eQ}{R})$

$E = 0$

6 0

3 years ago

A 5.00 kg mass is attached to a spring on a horizontal frictionless surface. The elastic constant of the spring is 37.9 N/m. If

Fynjy0 [20]

45.6 i think i'm not sure pls double check

6 0

3 years ago

Two loudspeakers placed 6.0 m apart are driven in phase by an audio oscillator, whose frequency range is 1595 to 2158 Hz. A poin

denpristay [2]

Answer:

The frequency produced by the oscillator, for which destructive interference occurs at point P, in the range of frequencies for the oscillator = 2033 Hz

Explanation:

For destructive interference, the difference in path length of the waves from each loudspeaker is related to the wavelength through.

(m + ½)λ = |d₁ - d₂|

where m can take on positive whole number values 0,1,2,3...

Point P is 4.7 m from A and 3.6 m from B

d₁ = 4.7 m

d₂ = 3.6 m

|d₁ - d₂| = 4.7 - 3.6 = 1.1 m

(m + ½)λ = |d₁ - d₂| = 1.1

(m + ½)λ = 1.1

where m could be any whole number from 0,1,2...

And the relationship between velocity of a wave, v, its frequency, f, and the wavelength, λ, is given as

v = fλ

The frequency range of the audio oscillator is frequency range is 1595 to 2158 Hz.

We can find a wavelength for the sound within this range, so as to obtain the exact frequency.

The options include 2001, 2033, 2127, 2095, or 2064 Hz.

Taking just 1 frequency in that range, f = 2033 Hz.

v = fλ

λ = (v/f) = (344/2033) = 0.169 m

Inserting in the destructive interference equation

(m + ½)λ = 1.1

If λ = 0.169 m

(m + ½) = (1.1/0.169) = 6.5

m + ½ = 6.5

Hence, it is evident that m = 6 for this question.

And the corresponding frequency at this level of destructive interference is 2033 Hz

Hope this Helps!!!

8 0

4 years ago