A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is
43.8 N = <em>k</em> (0.155 m) ==> <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m
The total work done on the spring to stretch it to 15.5 cm from equilibrium is
1/2 (283 N/m) (0.155 m)² ≈ 3.39 J
The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be
1/2 (283 N/m) (0.259 m)² ≈ 9.48 J
Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.
Answer:
You drive 0.025 miles blind (or 132 ft)
Explanation:
(distance) = (speed) x (time)
(distance blind) = (speed) x (time looking for CD)
(distance blind) = (45 mi/hr) x (2 s) x (1/3600 hr/s) = 0.025 miles = 132 ft
Answer:
quantity A is mass and quantity B is wright
