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3 years ago

A planet's density can be measured by combining

Physics

1 answer:

Lena [83]3 years ago

5 0

Answer:

a. Doppler and transit observations.

Explanation:

By means of the dopler effect, the radial velocity of an extrasolar planet can be measured, this is achieved by observing Doppler shifts in the spectrum of the star around which the planet orbits. This measure, combined with the transit observations, makes possible determine the density of the planet.

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Two spherical objects have masses of 3.1 x 10^5 kg and 6.5 x 10^3 kg. The gravitational attraction between them is 65 N. How far

nata0808 [166]

Answer:

4.55 x 10⁹m

Explanation:

Given parameters:

Mass of object 1 = 3.1 x 10⁵kg

Mass of object 2 = 6.5 x 10³kg

Gravitational force = 65N

Unknown:

Distance between them = ?

Solution:

To solve this problem, we use the expression below from the universal gravitational law;

Fg = $\frac{G mass 1 x mass 2}{distance ^{2} }$

G = 6.67 x 10⁻¹¹

65 = $\frac{6.67 x 10^{11} x 3.1 x 10^{5} x 6.5 x 10^{3} }{distance^{2} }$

Distance = 4.55 x 10⁹m

3 0

3 years ago

How does bilateral symmerty extend to our senses?

andreev551 [17]

Answer:

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8 0

3 years ago

A car follows a curved highway from point C to D, covering a distance of 100 kilometers and displacement to the east of 40 kilom

drek231 [11]

Answer:

40 speed and km/h

Explanation:

Velocity:40 km/h 2

8 0

3 years ago

A 1.89 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.77 N

Maksim231197 [3]

Answer:

Magnitude F(t)=26.6 N

Direction: -x

Explanation:

Given data

Spring constant K=4.77 N/m

Mass m=1.89 kg

Displace A=5.56m

Time t=3.96s

To find

Magnitude of force F

Solution

The angular frequency is given as

$w=\sqrt{\frac{K}{m} } \\w=\sqrt{\frac{4.77N/m}{1.89kg} }\\w=1.59rad/s$

Force on object is

$F(t)=-mAw^{2}Cos(wt)$

Substitute given values

$F(t)=-(1.89kg)(5.56m)(1.59rad/s)^{2}Cos(1.59*3.96)\\F(t)=-26.6N$

Magnitude F(t)=26.6 N

Direction: -x

4 0

3 years ago

A 280-g mass is mounted on a spring of constant k = 3.3 N/m. The damping constant for this damped system is b = 8.4 x 10^-3 kg/s

yulyashka [42]

Answer:

The number of oscillation is 36.

Explanation:

Given that

Mass = 280 g

Spring constant = 3.3 N/m

Damping constant $b=8.4\times10^{-3}\ Kg/s$

We need to check the system is under-damped, critical damped and over damped by comparing b with $2m\omega_{0}$

$2m\omega_{0}=2m\sqrt{\dfrac{k}{m}}$

$2\sqrt{km}=2\times\sqrt{3.3\times280\times10^{-3}}=1.92kg/s$

Here, b<< $2m\omega_{0}$

So, the motion is under-damped and will oscillate

$\omega=\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}$

The number of oscillation before the amplitude decays to $\dfrac{1}{e}$ of its original value

$A exp(\dfrac{-b}{2m}t)=A exp(-1)$

$\dfrac{b}{2m}t=1$

$t = \dfrac{2m}{b}$

$t=\dfrac{2\times280\times10^{-3}}{8.4\times10^{-3}}$

$t=66.67\ s$

We need to calculate the time period of one oscillation

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\times3.14}{\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}}$

$T=\dfrac{2\times3.14}{\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}}}$

$T=\dfrac{2\times3.14}{\sqrt{\dfrac{3.3}{280\times10^{-3}}-\dfrac{(8.4\times10^{-3})^2}{4\times(280\times10^{-3})^2}}}$

$T=1.83\ sec$

The number of oscillation is

$n=\dfrac{t}{T}$

$n=\dfrac{66.67}{1.83}$

$n=36$

Hence, The number of oscillation is 36.

8 0

3 years ago