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aleksandr82 [10.1K]
3 years ago
12

In 2009, NASA's Messenger spacecraft became the second spacecraft to orbit the planet Mercury. The spacecraft orbited at a heigh

t of 125 miles above Mercury's surface. You can assume this is a circular orbit. Determine the orbital speed and orbital period of Messenger. (GIVEN: RMercury= 2.44 x 106m; MMercury= 3.30 x 1023kg; 1 mi = 1609 m)
Physics
1 answer:
oee [108]3 years ago
5 0

Answer:

Explanation:

height, h = 125 miles = 125 x 1609 m = 201125 m = 0.2 x 10^6 m

Rdaius, R = 2.44 x 10^6 m

Mass, M = 3.3 x 10^23 kg

Let vo be the orbital speed.

The formula for the orbital speed is given by

v_{o}=\sqrt{\frac{GM}{R+h}}

where, M is the mass of mercury, R be the radius of mercury

v_{o}=\sqrt{\frac{6.67\times10^{-11}\3.3\times 10^{23}}{2.64\times 10^{6}}

vo = 2887.47 m/s

Time period, T = 2π(R+h) / vo

T = 2 x 3.14 x 2.64 x 10^6 / 2887.47

T = 5741.77 seconds

T = 1.6 hours

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valentina_108 [34]

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: v_f=v_0+a*t, where v_f is the final velocity, v_0 is the initial velocity, a the acceleration, and t is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s}  }{-201\frac{m}{s^2} }=1.40s, where v_f=0\frac{m}{s} because the sled is totally stopped, v_0=282\frac{m}{s} is the velocity of the sled before braking and, a=-201\frac{m}{s^2} is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration:x_f-x_0=v_0*t+\frac{1}{2}*a*t^2, where x_f-x_0 is the distance traveled, v_0 is the initial velocity, t the time of the process and, a is the acceleration of the process.

Then for this case the relationship becomes: x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m.

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Answer:

54 km/hr

Explanation:

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plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuu
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Answer:

12500 V

Explanation:

The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:

\Delta V = E d

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For the capacitor in this problem, we have

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Niobium wire with a 2.60 mm diameter has a maximum current capacity of 500 A while still remaining superconducting.

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Electron movement is referred to as electron current. The positive terminal receives electrons that are released by the negative terminal. Traditional current, usually referred to as just current, exhibits behavior consistent with positive charge carriers being the source of current flow. Regular current is received at the positive end and then flows to a negative terminal.

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