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Sati [7]
3 years ago
10

On July 15, 2004, NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected int

o an orbit 705 km above the earth's surface, and we shall assume a circular orbit.
(a) How many hours does it take this satellite to make one orbit?


(b) How fast (in km/s) is the Aura spacecraft moving?
Physics
1 answer:
Reika [66]3 years ago
7 0
On the earth surface g = GM/R^2
 And at the hieight h
 g" = GM/ [R+h] ^2
 g" = R^2/ [ R+h]^2 * g
 g" = {R/ [ R+h] }^2 * g
 g" = {6378/ 7083}^2 * 9.81 = 7.95 m/s^2
 --------------------------------------
T^2 = 4π^2 (R+h) /g"
T^2 = 4π^2 *7083 /7.95 
T = 187.54s = 0.052 hour.
 ========================
 Part B
 v = 2*π *[R+h] / T
 v = 2*π *[7.083] / 187.54s
 v = 0.24 km /s 
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seropon [69]

Answer:

1.884 meters per second

Explanation:

s = d/t

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t = 20 s

s = (37.68 m)/(20 s)

s = 1.884 m/s

7 0
2 years ago
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Answer:

Hey

Good question

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Hope it helped (i wrote all that).

Spiky Bob your answerer

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4 0
3 years ago
Two charged objects, A and B, are exerting an electric force on each other. What will happen if the charge on A is increased?
ozzi

Answer: The forces acting on both of them will increase in magnitude.

Explanation:

According to Coulomb's law, the electrostatic force between two bodies is proportional to the product of their two charges. If the charge on A is increased this product increases in size (it must have been non-zero to begin with, since there was a force between them at first). Thus, the force between them rises.

6 0
3 years ago
A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle
Sidana [21]

Answer:

B = 0.135T

Explanation:

To find the magnitude of the magnetic field you use the following formula, for the torque produced by a magnetic field B in a loop:

\tau=NIABsin\theta   (1)

τ: torque = 1.51*10^-5 Nm

I: current = 2.47mA = 2.47*10^-3 A

B: magnitude of the magnetic field

A: area of the loop = 4.97cm^2 = 4.97(10^-2m)^2=4.97*10^-4m^2

N: turns = 181

θ: angle between B and the magnetic dipole (same as the direction  of the normal to the plane)

You replace the values of the parameters in (1). Furthermore you do B the subject of the formula:

B=\frac{\tau}{NIAsin\tetha}=\frac{1.51*10^{-5}Nm}{(181)(2.47*10^{-3}A)(4.97*10^{-4}m^2)(sin30.1\°)}=0.135T

3 0
3 years ago
Using Ohms Law, calculate the current that would exist in a circuit having a voltage of 240V experiencing 110 Ohms of resistance
anygoal [31]

To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)

To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)

To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)

To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)

8 0
3 years ago
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