Explanation:
Given that,
Work done to stretch the spring, W = 130 J
Distance, x = 0.1 m
(a) We know that work done in stretching the spring is as follows :

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m
So,

So, the new work is more than 130 J.
Answer:
t = 13.7 s or t = 14 s with proper significant figures
Explanation:
The initial speed is 0 m/s since the car starts from rest, acceleration is 5.5 m/s2 and distance is 523 m.
Since we have initial speed, acceleration and distance we can use the following formula to find the time. We can now use algebra to work out our answer.
d = vt +
at²
523 = (0)t + (
)(5.5)t²
523 = 2.8t²
186.8 = t²
13.7 s = t
(t = 14 s with proper significant figures)
Answer:
S = V t where S is the horizontal distance traveled
1/2 g t^2 = H where H is the vertical distance traveled
t^2 = 2 H / g
V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations
tan theta = H / S
V^2 = S g / (2 tan theta)
Using S = L cos theta
V^2 = L g cos theta / (2 tan theta)
Giving V in terms of L and theta
The gravitational force on the car is
(9.8 m/s^2) x (the car's mass in kg).
The unit is newtons.
Answer:
A. Thickness and temperature
Explanation: