Asked by Anonymous
a 1000 kg car is traveling 27 m/s over 7 seconds what is it average forward force of the car?
27 m/s x 7 s =189 m/s^2 = a
1000 kg x 189 m/s^2 = 189,000 N
27 m/s x 7 s =189 m/s^2 = a
1000 kg x 189 m/s^2 = 189,000 N
Answers
Answered by
CodyJinks
v=at 27m/s=a(7s) a= 27/7= 3.86m/s^2 F=ma F= (1000kg)(3.86m/s^2) = 3860 N
FYI I am a physics student rn in college first semester of it. Bear with me if im wrong
FYI I am a physics student rn in college first semester of it. Bear with me if im wrong
Answered by
Damon
That question makes no sense as asked.
Ideally it take ZERO force to maintain a constant speed.
FORCE implies ACCELERATION
F = m A
Now perhaps you mean it accelerated from zero to 27m/s in 7 seconds
a = change in velocity / change in time
= 27/7 =3.86 m/s^2
F = 1000 kg * 3.857 = 3,857 Newtons
Ideally it take ZERO force to maintain a constant speed.
FORCE implies ACCELERATION
F = m A
Now perhaps you mean it accelerated from zero to 27m/s in 7 seconds
a = change in velocity / change in time
= 27/7 =3.86 m/s^2
F = 1000 kg * 3.857 = 3,857 Newtons
Answered by
Damon
Yes CodyJinks :)
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