Answer:
Explanation:
Geosynchronous orbit (GSO): Orbits with an altitude of approximately 35,786 km (22,236 mi). Such a satellite would trace an analemma (figure 8) in the sky. Geostationary orbit (GEO): A geosynchronous orbit with an inclination of zero.
The correct answer would be false. The less internal heat a jovian planet emits the lesser it stirs up its clouds making the atmosphere hotter. All of the four Jovian planets have unique atmospheres. They have more or less the same structures but they differ in their average temperature. As the distance of the planet is closer to the sun the atmosphere would be cooler. These planets are Jupiter, Saturn, Uranus and Neptune. They do not have a solid structure instead they are primarily composed of helium and hydrogen which makes them gas giants or ice planets. They are larger in size than the remaining planets in the solar system.
Answer: e = 3.333% or 33 1/3%
^
repeated
Identify the given information.
Work input (Wi)= 600 K
Work output (Wo) = 200 K
Choose which formula you should use. In this case, you are finding efficiency.
e = (Wo/Wi) x 100%
Substitute and solve.
e = (200 K/600 K) X 100%
e = 0.333 X 100%
e = 33.333% or 33 1/3%
False.
Unweathered rock that underlies soul is known as bedrock.
Answer:
3.67 N
Explanation:
From the question given above, the following data were obtained:
Charge of 1st object (q₁) = +15.5 μC
Charge of 2nd object (q₂) = –7.25 μC
Distance apart (r) = 0.525 m
Force (F) =?
Next, we shall convert micro coulomb (μC) to coulomb (C). This can be obtained as follow:
For the 1st object
1 μC = 1×10¯⁶ C
Therefore,
15.5 μC = 15.5 × 1×10¯⁶
15.5 μC = 15.5×10¯⁶ C
For the 2nd object:
1 μC = 1×10¯⁶ C
Therefore,
–7.25 μC = –7.25 × 1×10¯⁶
–7.25 μC = –7.25×10¯⁶ C
Finally, we shall determine the force. This can be obtained as follow:
Charge of 1st object (q₁) = +15.5×10¯⁶ C
Charge of 2nd object (q₂) = –7.25×10¯⁶ C
Distance apart (r) = 0.525 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 15.5×10¯⁶ × 7.25×10¯⁶ / 0.525²
F = 3.67 N
Therefore, the force on the object is 3.67 N
