Answer:
Waxing Gibbous
Third quarter
Waning Gibbous
Explanation:
If moon rises at 3:00 pm then the phase of the moon will be "Waxing Gibbous".
This is because, the moon is actually not fully illuminated but has achieved more than half of its full illumination.
If the moon is highest in the sky at sunrise then the phase of the Moon will be the "Third quarter"
This is because of the fact that at this position moon will rise at midnight, thus it will be at the highest point at the time of the sunrise.
If the moon sets at 10:00 am then the phase of the Moon is "Waning Gibbous"
This is because of the fact that at this position the Moon is moving towards becoming new Moon but at the same time, the moon is illuminated more than its half illumination.
-- Electric field lines DO never cross. <em>(A)
</em>
-- Electric field lines that are close together DO indicate a stronger electric field. <em>(B)
</em>
-- Electric field lines DO not affect the charge that created them. <em>(C)</em>
-- Electric field lines DON'T begin on north poles and end on south poles. North and South "poles" are the way we talk about magnets, not electric charges.
Answer:
The answer is "The object's speed relative to S can be greater than or less than its speed relative to S', depending on the actual values."
Explanation:
The S' frame and the object are moving in a positive direction. The object is moving with respect to the S frame so the S frame the rest frame
take the velocity of the object with respect to the rest frame as v and the velocity of the S' frame with respect S frame as v2
relative velocity of the object to the S' frame would be
Vrel = v2- v
This means the Vrel of the object with respect to the S' frame is less than the Vrel of the object with respect to the S frame
However is the S' velocity is greater than that of the object then the Vrel of the object with respect to the S' frame is greater than the Vrel of the object with respect to the S frame.
This would mean the second option is the answer, the relative speed of the object depends on the actual values.
1.1 A. An electric oven with a resistance of 201Ω and a voltage of 220V drwa a current of 1.1 A.
The easiest way to solve this problem is using the Ohm's Law I = V/R.
An electric oven has R = 201Ω, and a drop of voltage V = 220v, solve using I = V/R:
I = 220V / 201Ω
I = 1.09 A ≅ 1.1 A
