What processes formed the beaches of Florida?

A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou

kupik [55]

Answer:

The acceleration of the satellite is $0.87 m/s^{2}$

Explanation:

The acceleration in a circular motion is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

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

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite ( $1.50x10^{7} m$ ) and the Earth radius ( $6.38x10^{6} m$ ) :

$r = 1.50x10^{7} m+6.38x10^{6}m$

$r = 21.38x10^{6}m$

Then, equation 2 can be used:

$T = 8.65 hrs \cdot \frac{3600 s}{1hrs}$ ⇒ $31140 s$

$v = \frac{2 \pi (21.38x10^{6}m)}{31140s}$

$v = 4313 m/s$

Finally equation 1 can be used:

$a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}$

$a = 0.87 m/s^{2}$

Hence, the acceleration of the satellite is $0.87 m/s^{2}$

6 0

3 years ago

A lightning bolt with 13 kA strikes an object for 14 μ s. How much charge is deposited on the object?

pogonyaev

Answer:

0.182C

Explanation:

Using Q= It

= 13x10^3 . 14x10^-6

= 0.182C

8 0

3 years ago

A 5 kilograms bowling ball is dropped out a window. It hits the ground, and bounces upward. The velocity change of the ball is n

ioda

Answer:

13.5

Explanation:

Mass: 5kg

Initial Velocity: -15

Final Velocity: 12

Force: 10

We can use the equation: Vf = Vi + at

We need to find acceleration, and we can use the equation, F=ma,

We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.

Now we have all the variables to find time.

Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)

Plugging them in into desmos gives 13.5 for time.

4 0

2 years ago

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: