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natali 33 [55]
3 years ago
5

Which statement correctly describes the quantitative relationship between acceleration and net force on an object shown in the d

ata table? (1 point) When the net force applied to an object changes, the mass of the object remains constant. When the net force applied to an object changes, the acceleration changes by a different factor. When the net force applied to an object changes, the mass of the object also changes. When the net force applied to an object changes, the acceleration changes by the same factor.

Physics
1 answer:
Brrunno [24]3 years ago
7 0

Answer:

The correct answer is A

Explanation:

<h2><u>Because when the net force applied to an object changes, the acceleration changes by the same factor.  This answer is correct and I hope it helps you too.</u></h2><h2><u></u></h2>

Hope this helps....

Have a nice day!!!!

<h2><u></u></h2>
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Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
Alinara [238K]

Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

E=K\dfrac{q}{r^2}

E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

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