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Vika [28.1K]
3 years ago
13

.

Physics
1 answer:
Dmitry [639]3 years ago
5 0

Answer: 176.4 J

Explanation:

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A succession of single tones or pitches perceived as a unit is called a melody. It is a linear succesion of tones in music that is seen as one single entity. It is a combination of rhythm, pitch and tonal color. Melodies usually has one ormore motifs or phrases that are being repeated all throughout a composition. Melodies are characterized by the melodic motion it posses , the pitch, intervals between the pitches, continuity and pitch range. Due to the a wide range of melodies that can be made from tones, differet genres of music arise. Examples are Jazz, Rock music, Classical, pop, R&B and the like. 
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3 years ago
How is the electrical conductivity of a metal explained by metallic bonds.
love history [14]

Answer:

Metallic bonding is the bond that exist between the atoms. The electrons do not only flow at its respective atoms but instead, contribute to a sea of delocalised electrons. As a result, metals can conduct electricity as the delocalised electrons are able to carry charges.

Explanation:

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2 years ago
What is the second cooler layer of ocean water?
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4 years ago
Future space stations will create an artificial gravity by rotating. Consider a cylindrical space station 780 m diameter rotatin
ser-zykov [4K]

Answer:1.513 rps

Explanation:

Given

Diameter of cylindrical space d=780\ m

When the station rotates it creates centripetal acceleration which is given by

a_c=\omega ^2r

Now it must create the effect of gravity so

g=\omega ^2r

\omega =\sqrt{\frac{g}{r}}

\omega =0.158\ rad/s

and \omega =\frac{2\pi N}{60}

Thus N=\frac{4.755}{3.142}

N=1.513\ rps

5 0
3 years ago
Two charged concentric spherical shells have radii 8.83 cm and 15.4 cm. The charge on the inner shell is 5.03 × 10⁻⁸ C and that
REY [17]

Answer:

a. E(r) = 3.67 x 10⁴ N/C

b. E(r) = 5.38 x 10⁴ N/C

Explanation:

r1 = 8.83 cm

r2 = 15.4 cm

Charge on the inner shell = 5.03 x 10^-⁸ C

Charge on the outer shell = 2.77 x 10^-⁸ C

a)   To find the electric field at r = 11.1 cm,

r₁ = 8.83 cm < r = 11.1 cm < r₂ = 15.4 cm

E(r) = <u>1              </u> <u>q₁    </u>

          4\pi∈₀        r₂  

E(r) = <u>( 8.99 x 10⁹ N.m²/C² )  ( 5.03 x 10⁻⁸)</u>

                                 (0.111m)²

E(r) = 3.67 x 10⁴ N/C

b)   To find the electric field at r = 36.1 cm

since r₁  < r₂ < r = 36.1 cm

E(r) = <u>1              </u> <u>q₁  ₊  q₂   </u>

          4\pi∈₀           r₂  

E(r) = <u>( 8.99 x 10⁹ N.m²/C² )  ( 5.03 + 2.77 )(1 x 10⁻⁸)</u>

                                             (0.361m)²

E(r) = 5.38 x 10⁴ N/C

3 0
4 years ago
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