HELP ME PLEASE If an object is placed under a force of 20 N, it accelerates at a rate of 5 m/s^2. If the force is increased to 5
5 N, what is the new acceleration? a 13.75 kg b 7.25 kg c 21.10 kg d 5.65 kg
2 answers:
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Answer:
50.4 N
Explanation:
Q1 = Q
Q2 = 4 Q
Distance = d
The force is given by

.... (1)
Now,
Q3 = 2 Q
Q4 = 7 Q
distance = d/3

.... (2)
Divide equation (2) by equation (1), we get
F' / 1.60 = 126 / 4
F' = 50.4 N
Thus, the force is 50.4 N.
1,300 or more.
hope this helped :)
Hooke's Law says that F=-kx where k is the spring constant measured in N/m (newtons per meter)
The answer is in the attachment
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Answer:
C. 110 - 140
Explanation: