How much does horizontal velocity change each second as it moves through the air?

1 answer:

3 0

Well, Bianca, I'll tell ya. You really haven't told us enough to completely define exactly what's going on here. In fact, now that I look at it carefully, you've told us nothing at all ... only asked a question based on zero given information.

IF you're talking about an object like a baseball or a rock or another projectile of some kind, and IF the object has no jet engines on it, and IF it has been tossed, launched, thrown, or otherwise sent on its way to somewhere other than the place it left from, and IF it's OK to ignore the effects of air resistance, then ...

-- the only force acting on the object is the force of gravity

-- gravity only points down

-- gravity has no effect on horizontal velocity

-- horizontal velocity is <em>constant</em>, and <em>doesn't change </em>as time goes on.

But IF any of these things is different, then you can forget everything I said. I'm just guessing.

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In an atom that has an electron configuration of 1s^2 2s^2 wp^5, what is the total number of electrons in its sublevel of highes

frutty [35]

In an atom that has an electron configuration of 1s^2 2s^2 2p^3. The total number of electrons in its sublevel of highest energy is 3.

Electronic configuration represents the number of electrons in the excited state or in the subshells of an element.

1s^2 2s^2 2p^3 is the electronic configuration of nitrogen i.e. N7.

The atomic number of nitrogen is 7 and the atomic mass is 3. Hence, the electrons are in the 's' and 'p' subshells of the main shell.

This electronic configuration shows that nitrogen is in an exciting oxidation state and has 3 electrons in its highest energy level.

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8 0

2 years ago

Two tiny, spherical water drops, with identical charges of -7.08 × 10-16 C, have a center-to-center separation of 1.00 cm. (a) W

galben [10]

Answer:

(a) Electrostatic force F=4.51×10⁻¹⁷N

(b) Number of electron n=4425 electrons

Explanation:

Given data

Charges q₁=q₂= -7.08×10⁻¹⁶C

Distance r=1.00cm =0.01 m

To find

(a) Electrostatic force F

(b) Number of electron n

Solution

For (a) Electrostatic force

From Coulombs law we know that

$F=k\frac{q_{1}q_{2} }{r^{2} }\\ As\\ q_{1}=q_{2}=q\\So\\F=k\frac{q^{2}}{r^{2} }\\F=(8.99*10^{9})\frac{(-7.08*10)^{-16})^{2}}{(0.01)^{2} }\\F=4.51*10^{-17}N$