X = 1/2 a t square <br> Help me please

How did scientists discover the Earth had a liquid outer core and solid inner core?

viktelen [127]

Dr. Inge discovered the make up of the earths inner core by studying how an earthquakes waves bounced off the core. And Inge Lehmann was studying the waves of a 1929 earthquake when she found them acting inconsistently with solid mantle crust
hope it helps you

6 0

3 years ago

قوة الجذب المركزي تكون في اتجاه

pochemuha

Answer:

تكون دائمًا متعامدة مع سرعة الجسم وتكون دائمًا في اتجاه مركز انحناء المسار

Explanation:

6 0

3 years ago

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Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that:

$\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}$

$\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83$

6 0

3 years ago

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The acceleration reaches its minimal value of zero at the top of the trajectory. *

Serhud [2]

it is true. the trajectory reaches the value of zero at the top

4 0

3 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.83 times a second. A tack is stuck in the tire a

timama [110]

Answer:

The tangential speed of the tack is 6.988 meters per second.

Explanation:

The tangential speed experimented by the tack ( $v$ ), measured in meters per second, is equal to the product of the angular speed of the wheel ( $\omega$ ), measured in radians per second, and the distance of the tack respect to the rotation axis ( $R$ ), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second: