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3 years ago

Can anybody help me?

Physics

1 answer:

jonny [76]3 years ago

6 0

The answers have to be D and C

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Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

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Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

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3 years ago

Starting from rest, a 2.1x10^-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exert

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Answer:

a) The flea's speed when it leaves the ground is $v=1.88m/s$

b) The flea move $18cm$ upward while it is pushing off

Explanation:

<u>Knwons</u>

Mass $m=2.1\times 10^{-4} kg$ , Work $W=3.7\times 10^{-4} J$ and Force $F=0.41N$

a) Here we are going to use $W=\frac{1}{2} mv^{2}$ , so $v=\sqrt{\frac{2W}{m} }= \sqrt{\frac{2(3.7\times 10^{-4} J)}{2.1\times 10^{-4} kg} }=1.88m/s$

a) Here we are going to use $W=mgh$ , so $h=\frac{W}{mg}= \frac{3.7\times 10^{-4} J}{(2.1\times 10^{-4} kg)(9.8m/s^{2})} =0,1797m$ or $18cm$ approx.

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Answer: C

Explanation:

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HELP URGENT!!!!!!!!!!!!!!!!!!!!!!!!

nirvana33 [79]

D
Because the rest of the answers are illogical

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