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worty [1.4K]
3 years ago
7

Which Period 2 element would you expect to have the highest electrical and

Physics
1 answer:
evablogger [386]3 years ago
4 0
The answer is c nitrogen thx me later
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A raindrop of radius r falls from a certain height h above the ground. The work done by
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Answer:

fhc

Explanation:

chchfufuufufyfydyedhxhfud wyruficjc

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3 years ago
What is the approximate weight of a 20-kg cannonball on earth
laiz [17]
20 times 9.8 = 196 N
6 0
2 years ago
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A rocket travels at a speed of 14,000 m/s. What is the distance travelled by the rocket in 150s?
kati45 [8]

1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

v=\frac{S}{t}

Rearranging the equation, we can write

S=vt

In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is

S=vt=(14000 m/s)(150 s)=2.1 \cdot 10^6 m


2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:

S=vt=(14000 m/s)(300 s)=4.2 \cdot 10^6 m/s

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3 years ago
An airplane flies 1,592 miles east from Phoenix, Arizona, to Atlanta, Georgia, in 3.68 hours.
Gnoma [55]
Maybe it is around 300
8 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
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