Answer:
h=15.27m
Explanation:
Since at maximum height the vertical velocity must be null it's better to use the formula:
We will use this formula for the vertical direction, choosing the upward direction as the positive one, so we have:
or
which for our values is:
The rotation of Earth is equivalent to one day which is comprised of 24 hours. To determine the number of miles in Earth's circumference, one simply have to multiply the given rate by the appropriate conversion factor and dimensional analysis. This is shown below.
C = (1038 mi/h)(24 h/1 day)
C = 24,912 miles
From the given choices, the nearest value would have to be 20,000 mile. The answer is the second choice.
Answer:
Your answer would be letter <em><u>B</u></em><em><u>.</u></em><em><u> </u></em><em><u>Electrons</u></em><em><u> </u></em><em><u>orbit</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>energy</u></em><em><u> </u></em><em><u>level</u></em><em><u>.</u></em>
Explanation:
Hope it helps..
Just correct me if I'm wrong, okay?
But ur welcome!!
(;ŏ﹏ŏ)(◕ᴗ◕✿)
Answer:
q = 400 nC
the correct answer is b
Explanation:
The expression for the electric potential of a point charge is
V = k q / r
they ask us for the electrical charge
q = V r / k
let's calculate
Q = 600 6.0 / 9 10⁹
Q = 4 10⁻⁷ C
let's reduce to nC
Q = 4 10⁻⁷ C (10⁹ nC / 1C)
q = 4 10² nC = 400 nC
the correct answer is b
Traslate
La expresión para el potencial eléctrico de una carga puntual es
V = k q/r
nos piden la carga eléctrica
q= V r /k
calculemos
Q= 600 6,0 / 9 10⁹
Q= 4 10⁻⁷ C
reduzcamos a nC
Q = 4 10⁻⁷ C(10⁹ nC/1C )
q = 4 10² nC = 400 nC
la respuesta correcta es b
Answer: See below
Explanation:
The Earth attracts the falling object with the same intensity of gravity as the object attracts the Earth, according to Newton's law of gravitation. The displacement of the two bodies, however, is inversely proportional to their respective masses.
Example: The Earth attracts a ball that falls 3 metres from the ground, even though the ball's mass is insignificant in comparison to the Earth's. Similarly, the ball draws the Earth with the same power, but the Earth's mass is enormously more than the ball's. As a result, the Earth collides with a billionth of a millimetre ball (or even less). Restart the Earth's descent on the ball you'll never see again.
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| SHORTHAX |
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