In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass.
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were.
Ok, Rameses:
Elementary charge . . . . . 1.6 x 10⁻¹⁹ coulomb
negative on the electron
plussitive on the proton
Electron rest-mass . . . . . 9.11 x 10⁻³¹ kg
a). The force between two charges is
F = (9 x 10⁹) Q₁ Q₂ / R²
= (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²
= ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²
= 8.05 x 10⁻⁸ Newton .
b). Centripetal acceleration =
v² / r .
A = (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)
= 7.7 x 10²² m/s² .
That's an enormous acceleration ... about 7.85 x 10²¹ G's !
More than enough to cause the poor electron to lose its lunch.
It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use " F = MA "
either.
I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether F = M A .
I'm going to leave that step for you. Good luck !
A. “I usually get 10 piece nugget meal with medium fries at McDonald’s with a McFlurry
B. I could replace the 10 piece nuggets with a grilled chicken wrap and replace the medium fries with a small fries then replace the mcflurry with either water or unsweetened tea.
Answer:
The fence is 5feet less.
Explanation:
We need to determine
The less amount of fence required, if the enclosure has full width and reduced length, compared to full length and reduced width.
Approach & WorkingArea of lawn = 30 × 403/4th of the area of lawn = ¾(30 × 40) = 30 * 30
When full width will be fenced, and reduced length will be fenced.
Width = 30 feet30 * L = 30 * 30Hence, length = 30 feetLength of fence needed = 2(30 + 30) = 120 feet
When full length will be fenced, and reduced width will be fenced
Length = 40 feet40 * W = 30 * 30W = 22.5 feetLength of fence needed = 2(40 + 22.5) = 125 feet
Difference in length of fence needed = 125 – 120 = 5 feet.
Answer:
The correct answer is Dean has a period greater than San
Explanation:
Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.
T² = (4π² / G M) r³
When applying this equation to our case, the planet with a greater orbit must have a greater period.
Consequently Dean must have a period greater than San which has the smallest orbit
The correct answer is Dean has a period greater than San
