Answer:
<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg
k = 0.4
F = 200 N
<u>To </u><u>find </u><u>-</u><u> </u> acceleration
<u>Solution </u><u>-</u><u> </u>
F= kMA
200 = 0.4 * 20 * acceleration
200 = 8 * a
a = 8/200
a = 0.04 m s²
<h3>a = 0.04 m s²</h3>
to your question is 54 cm
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The new force would be 1.6 N.
Since the charges multiply as variables (q1)x(q2), then it would simply be double (q1)x(q2), or 1.6.
We know that
• The mass of the elevator is 5000 kg.
Let's draw a free-body diagram.
As you can observe, there are just two forces involved, the weight of the elevator and the tension force. Let's use Newton's Second Law.

But, W = mg = 5000kg*9.8m/s^2 = 49,000 N, and m = 5000 kg, a = 0 (because the speed is constant).

<h2>Therefore, the tension in the cable is 49,000 N.</h2>