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victus00 [196]
3 years ago
6

The mass of the sun is 1.9891030 kg and the mass of the Earth is 5.9721024kg. If the Earth’s acceleration toward the sun is 0.00

59 m/s/s, what is the sun’s acceleration toward the Earth?
Physics
1 answer:
saul85 [17]3 years ago
5 0

Answer:

this is the answer according to my calculations

Explanation:

0.001.9

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I believe if it were heavier with more mass, then the sun would pull it in and there would be no mercury. It might also be hotter.
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In which part of the scientific method do you tell why the research is being done?
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What is the proper way to start a fire?
Katyanochek1 [597]

Answer:

Explanation:

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3 years ago
A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa
Lesechka [4]
At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
U_i= \frac{1}{2} ka^2
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
K_i =0
And the total energy of the system is
E_i = U_i+K= \frac{1}{2}ka^2

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
U_f = 0
while the mass is moving at speed v, and therefore the kinetic energy is
K_f =  \frac{1}{2} mv^2
And the total energy is
E_f = U_f + K_f =  \frac{1}{2} mv^2

For the law of conservation of energy, the total energy must be conserved, therefore E_i = E_f. So we  can write
\frac{1}{2} ka^2 =  \frac{1}{2}mv^2
that we can solve to find an expression for v:
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6 0
3 years ago
A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl
USPshnik [31]

Given Information:  

Wavelength =  λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

λ = v/f

λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

L = 0.5596/2

L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

3 0
3 years ago
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