It looks like you have missed the necessary options for us to answer this question so I had to look for it. Anyway, here is the answer generally as the ℓ quantum number increases the energy is slightly higher. As a result 2s and 2p are no longer identical in energy in atoms<span> with more than one electron.</span>
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.
It is b. It gains momentum when going down. Think of it like a sled. When you go up the hill, it goes slower, but when you go down the hill, you gain speed(momentum) and it increases
<span>The initial momentum is 5 kg m/s and the final momentum is 10 kg m/s
'cause P = m*v, when we increase either m or v, P also increases by the same expression,
P = 5 * 2 = 10
So, option D is your answer!!
Hope this helps!
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<h2>Right answer: acceleration due to gravity is always the same </h2><h2 />
According to the experiments done and currently verified, in vacuum (this means there is not air or any fluid), all objects in free fall experience the same acceleration, which is <u>the acceleration of gravity</u>.
Now, in this case we are on Earth, so the gravity value is
Note the objects experience the acceleration of gravity regardless of their mass.
Nevertheless, on Earth we have air, hence <u>air resistance</u>, so the afirmation <em>"Free fall is a situation in which the only force acting upon an object is gravity" </em>is not completely true on Earth, unless the following condition is fulfiled:
If the air resistance is <u>too small</u> that we can approximate it to <u>zero</u> in the calculations, then in free fall the objects will accelerate downwards at
and hit the ground at approximately the same time.
Answer:
E = 20.03 J
Explanation:
Given that,
The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,
Voltage, V = 12 V
We need to find the energy delivered to the lightbulb filament during 2.00 s.
The energy delivered is given by :
. ....(1)
As,

As per Ohm's law, V = IR

Using formula (1).

So, the energy delivered to the lightbulb filament is 20.03 J.