determine the acceleration that result when a 12-N net force is applied to a 3 kg object and to a 6 kg object.
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Answer:
) By the definition of the Newton's Second Law of Motion, we have:
F_{net}=ma,F
net
=ma,
a=\dfrac{F_{net}}{m}=\dfrac{12\ N}{3\ kg}=4\ \dfrac{m}{s^2}.a=
m
F
net
=
3 kg
12 N
=4
s
2
m
.
b) We can find the acceleration of the 6 kg object using the same formula:
a=\dfrac{F_{net}}{m}=\dfrac{12\ N}{6\ kg}=2\ \dfrac{m}{s^2}.a=
m
F
net
=
6 kg
12 N
=2
s
2
m
.
Answer:
a) a=4\ \dfrac{m}{s^2}.a=4
s
2
m
.
b) a=2\ \dfrac{m}{s^2}.a=2
s
2
m
.
Explanation:
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