Answer:
Explanation:
Normal length of spring = 28.3 cm
stretched length of spring = 38.2 cm
length of extension = 38.2 - 28.3 = 9.9 cm
= 9.9 x 10⁻² m
force applied to stretch = .55 x 9.8 ( mg )
= 5.39 N
Force constant = force applied / extension
= 5.39 / 9.9 x 10⁻²
= .5444 x 10² N /m
= 54.44 N/m
Answer: 13.2 seconds.
Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0
S= distance travelled
a = acceleration due gravity
t= time.
1 foot = 0.305m so,
S= 2860 feet =872.3m
S= ut+1/2 at²
872.3 = 0×t + 1/2×10 × t²
872.3 =0 + 5t²
T²= 872.3/5
T²= 174.46
Take the square root of T we then have;
t = 13.2 seconds to one decimal place.
Since car is initially at rest so at that position the net force on car in vertical as well as horizontal both direction is zero
but after lights are green the car will speed up in horizontal direction
so now we can say that
It will accelerate in horizontal direction
So now as per Newton's formula we have

so now the force in horizontal direction will become nonzero while in vertical direction car is still in equilibrium
So now we can say


so the correct answer will be
C. The net force on the car is greater than zero in the horizontal direction.