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Veronika [31]
3 years ago
11

A point charge of +3.0 X 10-7 coulomb is placed 2.0 X 10-2 meter from a second point charge of +4.0 X 10-7 coulomb. What is the

magnitude of the electrostatic force on the charges?
A. 6.0 X 10-12 N
B. 3.0 X 10-10 N
C. 5.4 X 10-2 N
D. 2.7 N
Physics
1 answer:
Mars2501 [29]3 years ago
7 0

Answer:

D. 2.7 N

Explanation:

Applying

F = kq'q/r²................ Equation 1

Where F = force, k = coulomb's constant, q' = first charge, q = second charge, r = distance between the charge

From the question,

Given: q' = +3.0×10⁻⁷ C, q = +4.0×10⁻⁷C, r = 2.0×10⁻² m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation 1

F = ( +3.0×10⁻⁷)(+4.0×10⁻⁷)(8.98×10⁹)/(2.0×10⁻²)²

F = 26.94×10⁻¹ N

F = 2.694 N

F ≈ 2.7 N

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Imagine an alternative universe where the characteristic decay time of neutrons is 3 min instead of 15 min. All other properties
FrozenT [24]

Answer:

(a) [Y_{p} ]_{max} = \frac{2f}{1+f}

(b) f_{new} = 0.013; [Y_{p} ]_{max} = 0.026

Explanation:

Since the neutron-to-proton ratio at the time of nucleosynthesis is given:

f = \frac{n_{n} }{n_{p} }

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n_{n} = f*n_{p}

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Assuming that m_{p}=m_{n}, there would be a total of (n_{n}+n_{p}) protons and neutrons with a total mass of (n_{n}+n_{p})*m_{p}.

Thus:[Y_{p} ]_{max} = \frac{2f}{1+f}

(b) Given:

t_{nuc} = 200 s;   τ_{n} = 3*60s = 180 s

f_{new} = \frac{n_{nf} }{n_{pf} } = \frac{exp (-200/180)}{5 +[1- exp(-200/180)]} =\frac{0.077}{5.923} = 0.013

[Y_{p} ]_{max} = \frac{2f}{1+f} = (2*0.013)/(1+0.013) = 0.026

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3 years ago
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