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podryga [215]
3 years ago
11

1.

Physics
2 answers:
Firdavs [7]3 years ago
4 0

1). The forces inside the atom are always, totally, completely, electrostatic forces. Those are so awesomely stronger than the gravitational forces that the gravitational ones are totally ignored, and it doesn't change a thing.

Parts 2 and 3 of this question are here to show us how the forces compare.

Part-2). The electrostatic force between a proton and an electron.

The constant in the formula is 9x10^9, and the elementary charge is 1.602 x 10^-19 Coulomb ... same charge on both particles, but opposite signs.

I worked through it 3 times and got 0.000105 N every time. So the best choice is 'C', even though we disagree by a factor of ten times. You'll see in part-3 that it really doesn't make any difference.

Part-3). Gravitational force between a proton and an electron.

The constant in Newton's gravity formula is 6.67x10^-11 . You'll have to look up the masses of the proton and the electron.

I got 2.163 x 10^-55 N ... exactly choice-C. yay !

Now, after we've slaved over a hot calculator all night, the thing that really amazes us is not only that the electrostatic force is stronger than the gravitational force, but HOW MUCH stronger ... 10^51 TIMES stronger. That's a thousand trillion trillion trillion trillion times stronger !

That's why it has no effect on the measurements if we just forget all about the gravitational forces inside the atom.

bearhunter [10]3 years ago
4 0

Answer:

The correct answer is Fel > Fg

Explanation:

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A student weighing 120 lbs climbs a 12 ft flight of stairs in 9 seconds. how much power did the student create?
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It should be noted that velocity is calculated by dividing displacement by time. Thus, from the given in this item we can calculate for the power. 
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The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo
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And since K=7.35 J, we can find the velocity, v:
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3 0
3 years ago
A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0
3 years ago
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