The earth and moon are seperated by a

Physics

1 answer:

cestrela7 [59]3 years ago

7 0

Answer:

4674568 m or 4675 km

Explanation:

Let x be the distance from the center of the mass system to the center of the earth. This means the distance from the center of the mass system to the center of the moon is $3.85*10^8 - x$ . For the masses to balance out, we have the following equation:

$m_e*x = m_m*(3.85*10^8 - x)$

where $m_e$ is the mass of Earth, and $m_m$ is mass of the moon

$5.98 * 10^{24}x = 7.35*10^{22}(3.85*10^8 - x)$

We can divide both sides by $10^{22}$

$598x = 7.35*3.85*10^8 - 7.35x$

$605.35x = 2829750000$

$x = 2829750000/605.35 \approx 4674568 m$ or 4675 km

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A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m

ArbitrLikvidat [17]

Answer:

Explanation:

The sum of force along both components x and y is expressed as;

$\sum Fx = ma_x \ and \ \sum Fy = ma_y$

The magnitude of the net force which is also known as the resultant will be expressed as $R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}$

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

$a_x = \frac{d^2 x }{dt^2}$

$a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}$

Similarly,

$a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2$