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bogdanovich [222]
3 years ago
14

Student C said, “I don’t agree with either model you drew. I think the Sun goes around Earth. The evidence is that you can see t

he Sun rise in the east every morning, travel across the sky during the day, and set in the west every evening.”
Explain to Student C how a different frame of reference is better to explain why day and night occur.
Physics
1 answer:
xxTIMURxx [149]3 years ago
5 0

Answer:

See the explanation below

Explanation:

Student c's belief is fulfilled only for the movement of the earth with respect to the sun. But it has no validity or does not exist when it is necessary to explain other physical phenomena with respect to other satellites. For example, how would you explain the phases of the moon, if the Earth is located in the center of the Galaxy?, another serious question regarding the observations made by scientists millions of years ago, where they observed that the distances of the Earth from other planets were changing, with respect to time. If the Earth was in the center, all the planets and the sun would revolve around it preserving a constant distance (radius), at all times.

Other phenomena to explain would be the seasons on Earth, these are due to the axis of inclination of the Earth and the rotation of it around the sun.

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A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp
amm1812

Answer:

the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = -\beta\frac{dx}{dt} = \frac{-10dx}{dt}

\beta = 10

By Newtons second law ,

The diffrential equation of motion with damping is given by

m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}

substitute the value of m =1kg, k = 21N/M, and \beta = 10

1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}

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suppose the equation of the form x =e^m^t,

and the auxilliary equation is given by

m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3

The general solution for the above differential equation is

x(t) =C_1e^{-3t}+C_2e^{-7t}

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(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

x'(0) = 0

substitute the initial condition

C_1 +C_2 = 1

-3C_1-7C_2=0

Solve the above equation to get C₁ and C₂

C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}

substitute for C₁ and C₂ in general solution

x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

Thus the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

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