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Luda [366]
3 years ago
9

A proton is fired from very far away directly at a fixed particle with charge q = 1.82 ✕ 10−18 C. If the initial speed of the pr

oton is 3.5 ✕ 105 m/s, what is its distance of closest approach to the fixed particle? The mass of a proton is 1.67 ✕ 10−27 kg.
Physics
1 answer:
Virty [35]3 years ago
8 0

Answer:

   r = 2.56 10⁻¹² m

Explanation:

For this exercise let's use energy conservation

Starting point, proton too far

             Em₀ = K = ½ m v²

Final point. When proton is stop

             Emf = U = k q₁ q₂ / r

How energy is conserved

               Em₀ = Emf

              ½ m v² = k q₁ q₂ / r

              r = 2k q₁ q₂  / m v²

Let's calculate

             r = 2 8.99 10⁸ 1.6 10⁻¹⁹ 1.82 10⁻¹⁸ / (1.67 10⁻²⁷ (3.5 10⁵)² )

             r = 2.56 10⁻¹² m

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From shortest wavelength to longest wavelength:

Gamma Rays

X-Rays

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Visible Light

Infrared waves

Microwaves

Radio Waves

Explanation:

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elena55 [62]

The correct answer D: all of the above

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A 290 gg bird flying along at 6.2 m/sm/s sees a 9.0 gg insect heading straight toward it with a speed of 34 m/sm/s (as measured
Murrr4er [49]

Answer:

The bird's speed immediately after swallowing is 4.98 m/s.

Explanation:

Given that,

Mass of bird = 290 g

Speed = 6.2 m/s

Mass of sees = 9.0 g

Speed = 34 m/s

We need to calculate the bird's speed immediately after swallowing

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3}

Put the value into the formula

0.290\times6.2+0.009\times(-34)=(0.290+0.009)\times v_{3}

v_{3}=\dfrac{0.290\times6.2-0.009\times34}{(0.290+0.009)}

v_{3}=4.98\ m/s

Hence, The bird's speed immediately after swallowing is 4.98 m/s.

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(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of
NISA [10]

Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

F = \dfrac{kq_1q_2}{r^2}

where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.

8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}

Force due to Q₃ :

F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}

8.3. The net force on the particle at Q₂ is the vector

\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}

Its magnitude is

\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}

where we subtract 180° because \vec F terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.

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