What is the term for an area of a country located away from capital?

Please solve for 15 points. Please don’t input a link.

melomori [17]

Answer:

a). Single replacement.

Explanation:

Because one element replaces another element in a compound

6 0

3 years ago

Match the property of a sound wave to the correct definition. A. amplitude . B. frequency . C. period . D. pitch E. loudness

Maru [420]

Amplitude: How dense the medium is in the compression part of the wave, and how empty the rarefied area is.

Frequency: The number of wavelengths that pass a position in 1 second.

loudness: The quality of the sound that is most closely linked to the amplitude of the sound wave.

Period: The amount of time that it takes one wavelength to pass by a position.

Pitch: The quality of the sound that is most closely linked to the frequency of the sound wave.

5 0

3 years ago

Read 2 more answers

There is a puncture in the inner tor of your bicycle. Give a method to detect the position of the puncture

Pani-rosa [81]

Answer:

Explanation:

The first method to engage is to listen to where the sound of air in the inner Tor escaping originated and look to see if u can find it. You can then feel the escape air with your hand.

You can Put it inside a container of water and see the bubble and rotate the inner tube to pass all of it through the water

7 0

3 years ago

Atoms that are bonded together to form a new material with new properties and characteristics. .

mihalych1998 [28]

Answer:

Explanation:

Atoms form chemical bonds to make their outer electron shells more stable. ... An ionic bond, where one atom essentially donates an electron to another, forms when one atom becomes stable by losing its outer electrons and the other atoms become stable (usually by filling its valence shell) by gaining the electrons.

7 0

3 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

The current I is any motion of charge from one region to another, so this is given by:

 $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$

Finally, for the drift velocity magnitude vd, we find:

 $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

3 years ago