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3 years ago

How are scientists able to measure the age of earth?

Physics

1 answer:

RoseWind [281]3 years ago

4 0

Explanation:

Through the uranium-lead dating, you can establish the age of rocks that formed from 1 million years ago to 4.5 billion years with enough precision. Dating meteorites using this method, It is possible to establish the age of the earth, since these are pieces that are present since the formation of the solar system, so they are approximately the same age as the Earth.

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A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

$A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}$

Differentiate both sides with respect to d.

$\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)$

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

$\frac{d^{2}A}{dt^{2}}= + ve$

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

8 0

3 years ago

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

Circular velocity for the particle in the Encke Division:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

Circular velocity for the particle in D Ring:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$