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Rudiy27
4 years ago
15

A proton at location A makes an electricfield E-->1 atlocation B. A different proton, placed atlocation B, experiences a forc

e F--->1.
If |E-->|= 300 N/C, what is |F-->1|?
|F-->1|= N
Now the proton at B is removed and replaced by alithium nucleus, containing three protons and four neutrons. Theproton at location A remains in place.
What is the magnitude of the electric force on the lithium nucleus?
|F-->Li| = N
Now the Lithium nucleus is removed, and an electron is placed atlocation B. The proton atlocation A remains in place. What is the magnitude of the electric force on the electron?
|F-->1| = N
Physics
1 answer:
luda_lava [24]4 years ago
7 0

Answer:

Explanation:

 |F-->1| =  E-->1 x charge on proton

= 300 x 1.6 x 10⁻¹⁹

Magnitude of electric force on proton

|F-->1| = 480 x 10⁻¹⁹ N .

Inside lithium nucleus , there are three protons so charge on it

= 3 x 1.6 x 10⁻¹⁹ = 4.8 x 10⁻¹⁹ C .

Magnitude of electric force on lithium nucleus

|F-->Li|  =   E-->1 x charge on lithium nucleus

= 300 x  4.8 x 10⁻¹⁹

= 1440 x 10⁻¹⁹ N .

Electron has same charge as that possessed by proton so charge on electron = 1.6 x 10⁻¹⁹ C

Magnitude of force on electron

|F-->1| =  E-->1 x charge on electron

= 300 x 1.6 x 10⁻¹⁹

=  480 x 10⁻¹⁹ N .

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Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

     cos ( 360-30) = cos (-30) = x₁ / d

     sin (-30) = y₁ / d

     x₁ = d cos (-30)

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     x₁ = 1.2 cos (-30) = 1,039 miles

     y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

       cos (270-20) = x₂ / d

       cos (250) = y₂ / d

       x₂ = 2.0 cos 250 = -0.684 miles

       y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

       cos (180 + 30) = x₃ / d

       sin (210) = y₃ / d

       x₃ = 1.6 cos 210 = -1.3856 miles

       y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

       cos (90 + 15) = x₄ / d

       sin (105) = y₄ / d

       x₄ = 2.6 cos 105 = -0.6729 miles

       y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis  (West-East direction)

       X = x₁ + x₂ + x₃ + x₄

       X = 1.039 -0.684 - 1.3846 - 0.6729

       X = -1.7025 miles

   

       Y = y₁ + y₂ + y₃ + y₄

       Y = -0.6 -1.879 -0.8 +2.511

       Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

      D = √ (X² + Y²)

      D = √ (1.7025² + 0.768²)

      D = 1.8677 miles

let's use trigonometry to find the direction

       tan θ = Y / X

       θ = tan⁻¹ Y / x

       θ = tan⁻¹ (0.768 / 1.7025)

       θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

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Substituting the numbers into the formula, we find

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2) 5 N

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- The force of friction, which acts in the opposite direction (so, towards the left), of magnitude F_f = 5 N

Since the two forces are in opposite directions, the net force is given by their difference:

F_{net}=F-F_f = 10 N-5 N=5 N


3) Yes

The crate is accelerating. In fact, according to Second Newton's Law:

F_{net}=ma (1)

where Fnet is the net force on the crate, m is its mass, a is its acceleration. We can immediately see that since Fnet is not zero, the acceleration is also non-zero, so the crate is accelerating.

We can even calculate the magnitude of the acceleration. In fact, the mass of the crate is given by:

m=\frac{Weight}{g}=\frac{25 N}{9.8 m/s^2}=2.55 kg

And by using (1) we find

a=\frac{F_{net}}{m}=\frac{5 N}{2.55 kg}=1.96 m/s^2


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