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3 years ago

11

As an electron approaches a proton, the electric force of attraction

2 answers:

Andreas93 [3]3 years ago

4 0

I believe the answer is option B. As an electron goes near a proton, the force of attraction would increase. Electron is negatively charged while the proton is positively charged so they would naturally attract each other. Distance and force of attraction is inversely related so that as the distance decreases, the force increases.

Ilya [14]3 years ago

3 0

Answer:

B

Explanation:

Electromagnetism pulls the two together so the force becomes bigger

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4 years ago

Read 2 more answers

A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.90 kg box that is sitting on the horizontal,

Lady_Fox [76]

Answer:

6.875 m/s

Explanation:

The force is variable which is given by

F(x) = 18 - 0.53 x

mass of the box, m = 8.9 kg

initially it is at rest at x = 0

Let the velocity is v after travelling a distance of 15 m.

According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the body.

Work done = change in kinetic energy

$\int \overrightarrow{F}.d\overrightarrow{x}=\Delta K.E$

$\int_{0}^{15} \left ( 18-0.53 x \right )dx=\frac{1}{2}\times m \left ( v^{2}-u^{2} \right )$

$\left ( 18x-0.265x^{2} \right )_{0}^{15}=\frac{1}{2}\times 8.9\times \left ( v^{2}-0^{2} \right )$

18 x 15 - 0.265 x 15 x 15 = 4.45 x v²

270 - 59.625 = 4.45 v²

v² = 47.275

v = 6.875 m/s

Thus, the final velocity of the box is 6.875 m/s.

4 0

3 years ago

A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow

shusha [124]

Answer:

150000000

$\dfrac{F_e}{F_g}=0.00000203873598369$

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

$1\ C=6.25\times 10^{18}\ electrons$

Number electrons is

$n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000$

The number of electrons added or removed was 150000000

Force is given by

$F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N$

The ratio is

$\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369$

The ratio is $\dfrac{F_e}{F_g}=0.00000203873598369$

Balancing the forces we get

$Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C$

The electric field required is 49050000 N/C

4 0

4 years ago