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melomori [17]
2 years ago
6

An electron and a proton are separated by a distance of 1.6×10−10m

Physics
1 answer:
lesya [120]2 years ago
4 0

Answer:

Fe = 9 x 10⁻⁹ N

Fg = 3.97 x 10⁻⁴⁶ N

Fe = 2.26 x 10³⁷ Fg

Explanation:

First we find electric force by Coulomb's Law as follows:

Fe = kq₁q₂/r²

where,

Fe = electric force = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = q₂ = charges on electron and proton = 1.6 x 10⁻¹⁹ C

r = distance between electron and proton = 1.6 x 10⁻¹⁰ m

Therefore,

Fe = (9 x 10⁹ N.m²/C²)(1.6 x 10⁻¹⁹ C)(1.6 x 10⁻¹⁹ C)/(1.6 x 10⁻¹⁰ m)²

<u>Fe = 9 x 10⁻⁹ N </u>

Now we find gravitational force by Newton's Law of Gravitation as follows:

Fg = Gm₁m₂/r²

where,

Fg = gravitational force = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of electron = 9.11 x 10⁻³¹ kg

m₂ = mass of proton = 1.673 x 10⁻²⁷ kg

r = distance between electron and proton = 1.6 x 10⁻¹⁰ m

Therefore,

Fg = (6.67 x 10⁻¹¹ N.m²/kg²)(9.11 x 10⁻³¹ kg)(1.673 x 10⁻²⁷ kg)/(1.6 x 10⁻¹⁰ m)²

<u>Fg = 3.97 x 10⁻⁴⁶ N</u>

Dividing both forces:

Fe/Fg = (9 x 10⁻⁹ N)/(3.97 x 10⁻⁴⁶ N)

<u>Fe = 2.26 x 10³⁷ Fg</u>

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A car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began
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v = v0 + a t

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A train traveling at 25 m/s is blowing its whistle at 440 Hz as it crosses a level crossing. You are waiting at the crossing and
ohaa [14]

Answer:

b) 472HZ, 408HZ

Explanation:

To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:

f_o=f\frac{v_s+v_o}{v_s-v}\\\\f_o=f'\frac{v_s-v_o}{v_s+v}\\\\

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vs: speed of sound = 343m/s

vo: speed of the observer = 0m/s (at rest)

v: sped of the train

f: frequency perceived when the train leaves us.

f': frequency when the train is getTing closer.

Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:

f=f_o\frac{v_s-v}{v_s+v_o}=(440Hz)\frac{340m/s-25m/s}{340m/s}=408Hz\\\\f'=f_o\frac{v_s+v}{v_s-v_o}=(440Hz)\frac{340m/s+25m/s}{340m/s}=472Hz

hence, the frequencies for before and after tha train has past are

b) 472HZ, 408HZ

6 0
3 years ago
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