A )
t 1 = 2 h, t 2 = 6 h
Δ t = t 2 - t 1 = 6 - 2 = 4 h
54 = 50 + a Δ t
54 = 50 + 4 a
4 a = 54 - 4
4 a = 4
a = 4 : 4
a = 1 km/h²
v o = 48 km/h
An equation that can be used to describe the velocity of the car at the different times is:
v = 48 + t
B ) The graph is in the attachment.
It can be, but set also means many other things.
Answer:
271.862 N/m
Explanation:
From Hook's Law,
mgh = 1/2ke²............... Equation 1
Where
m = mass of the ball, g = acceleration due to gravity, k = spring constant, e = extension, h = height fro which the ball was dropped.
Making k the subject of the equation,
k =2mgh/k²....................... Equation 2
Note: The potential energy of the ball is equal to the elastic potential energy of the spring.
Given: m = 60.3 g = 0.0603 kg, g = 9.8 m/s², e = 4.68317 cm = 0.0468317 m, h = 53.7 cm = 0.537 m
Substitute into equation 2
k = 2(0.0603)(9.8)(0.537)/0.048317²
k = 0.6346696/0.0023345
k = 271.862 N/m
Answer:
Explanation:
Although there is absolutely NO regard for significant digits, I can help you with this, nonetheless.
The equation for Potential Energy is PE = mgh. We have everything but the height of the ball. We have to solve for that using a one-dimensional motion equation:
v² = v₀² + 2aΔx, where Δx is our displacement (the height we need for PE). Filling in and keeping in mind that at the max height of parabolic travel, the final velocity of the object is 0:
0 = (21.5)² + 2(-9.8)Δx and
0 = 462.25 - 19.6Δx and
-462.25 = -19.6Δx so
Δx = 23.58 m. Using this as the h in our PE equation:
PE = .19(9.8)(23.58) so
PE = 43.9 J, choice C.
