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aliya0001 [1]
3 years ago
13

Which statement is correct? When a positively charged atom looses an electron to a positively charged atom, two neutral atoms ar

e created. When a neutral atom looses an electron to another neutral atom, two charged atoms are created. When a neutral atom looses an electron to a positively charged atom, both atoms become neutral. When charged atoms lose electrons they will become positively charged atoms.
Physics
2 answers:
Setler79 [48]3 years ago
7 0

Answer:

When a neutral atom looses an electron to another neutral atom, two charged atoms are created.

Explanation:

Initially, we have two neutral atoms: they both have electrical charge equal to zero.

If one of the two neutral atom looses an electron, it becomes positively charge (charge: +1), because the electron carries a charge of -1. As a result, the other atom which accepts the electron becomes negatively charged (charge: -1). Therefore, in the end we will have two charged atoms.

vazorg [7]3 years ago
3 0
I believe the answer is "When a neutral atom looses an electron to another neutral atom, two charged atoms are created."
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The maximum velocity of the rock is 64.55 m/s

Answer: Option B

<u>Explanation:</u>

The possessed energy by the objects while motion called kinetic energy. It is directly proportionate to mass and square of velocity. The stored energy in the system called potential energy and expressed as below,

\text { potential energy }=\frac{1}{2} \times(\text { spring constant }) \times\left(\text { distance from equilibrium) }^{2}\right.

                  U=\frac{1}{2} \times k \times x^{2}

U = spring’s potential energy in certain place

k = the spring constant, in N/m.

x = the spring’s distance is stretched or compressed away from equilibrium

Conservation of energy states energy neither created nor destroyed but change from one form to other. So, using this concept, equating both potential and kinetic energies equation we get,

            \frac{1}{2} \times m \times v^{2}=\frac{1}{2} \times k \times x^{2}

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m = 12 g = 0.012 kg

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x = 0.500 m

Substitute these values we get,

         \frac{1}{2} \times 0.012 \times v^{2}=\frac{1}{2} \times 200 \times(0.500)^{2}

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7 0
3 years ago
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xmax = 9.5cm

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m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

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You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

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