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slega [8]
3 years ago
15

Help me it ask each planet distance from the sun.

Physics
1 answer:
raketka [301]3 years ago
4 0

Answer:

A Jupiter

B Neptune

C Saturn

D Uranus

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A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s
sergey [27]

Answer:

Y = V / f      where Y equals wavelength

4 Y1 = V / f1       for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2   for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1      dividing equations

f2 = 2 f1  

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

8 0
3 years ago
O operate this generator, a human applies a force by turning the crank. This action generates an electric current, which lights
Shtirlitz [24]
Answer A I’m very positive about it
3 0
3 years ago
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We know velocity of sound v=a₩(omega), but speed does not depend on amplitude, why?
MrRissso [65]
Hope it helps............

8 0
3 years ago
Two planets, Dean and Sam, orbit the Sun. They each have with circular orbits, but orbit at different distances from the Sun. De
lyudmila [28]

Answer:

The correct answer is Dean has a period greater than San

Explanation:

Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.

                T² = (4π² / G M)  r³

When applying this equation to our case, the planet with a greater orbit must have a greater period.

Consequently Dean must have a period greater than San which has the smallest orbit

The correct answer is Dean has a period greater than San

3 0
3 years ago
Assume that the radius ????r of a sphere is expanding at a rate of 70 cm/min.70 cm/min. The volume of a sphere is ????=43???????
rodikova [14]

Answer:

the rate of change in volume with time is 280πr² cm³/min

Explanation:

Data provided in the question:

Radius of the sphere as 'r'

\frac{d\textup{r}}{\textup{dt}}  = 70 cm/min

Volume of the sphere, V = \frac{\textup{4}}{\textup{3}}\pi r^3

Surface area of the sphere as 4πr²

Now,

Rate of change in volume with time, \frac{d\textup{V}}{\textup{dt}}

 = \frac{d(\frac{\textup{4}}{\textup{3}}\pi r^3)}{dt}

= 3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times\frac{dr}{dt}

Substituting the value of \frac{dr}{dt}

= 3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times70

= 280πr² cm³/min

Hence, the rate of change in volume with time is 280πr² cm³/min

4 0
3 years ago
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