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2 years ago

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The protons in a nucleus are approximately 2 ✕ 10^−15 m apart. Consider the case where the protons are a distance d = 1.93 ✕ 10^

−15 m apart. Calculate the magnitude of the electric force (in N) between two protons at this distance.

2 answers:

Ad libitum [116K]2 years ago

8 0

Answer:

61.8 N

Explanation:

Given data

Charge of the protons (q): 1.60 × 10⁻¹⁹ Coulomb

Distance between the protons (d): 1.93 × 10⁻¹⁵ meters

Coulomb's constant (k): 8.99 × 10⁹ N.m².C⁻²

We can find the magnitude of the electric force (F) between the two protons using Coulomb's law.

$F=k.\frac{q_{1}q_{2}}{d^{2} } \\F=8.99 \times 10^{9} N.m^{2} .C^{-2} .\frac{(1.60 \times 10^{-19}C)^{2} }{(1.93 \times 10^{-15}m)^{2} } \\F=61.8N$

The magnitude of the electric force is 61.8 N.

Bond [772]2 years ago

6 0

Answer: 61.2N

F=kq1q2/r^2

F=8.98E9(1.6E-19)(1.6E-19)/(1.93E-15)^2

F=61.2N

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There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

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Part B: v_1 = v_2 = $\frac{v}{2}$

Part C: v_1 = $\frac{v}{3}$ ; v_2 = $\frac{4v}{3}$

Part D: v_1 = v_2 = $\frac{v}{4}$

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To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

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v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

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<u>Part B</u>: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

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v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.2m}{2m+m}.v$

v_2 = $\frac{4v}{3}$

<u>Part D</u>: Object 1 is m, object is 3m and collision is inelastic:

v_1 = v_2 = $v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m}{m+3m}.v$

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