Answer:
a) 19.2 s
b) No
Explanation:
Given:
v₀ = 125 m/s
a = -6.5 m/s²
v = 0 m/s
a) Find: t
v = at + v₀
(0 m/s) = (-6.5 m/s²) t + (125 m/s)
t ≈ 19.2 s
b) Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (125 m/s)² + 2 (-6.5 m/s²) Δx
Δx ≈ 1200 m
An aircraft carrier that's 850 meters long won't be long enough.
Answer:
where is the graph I can't see it how can I solve the problem if I don't see the graph can you show the graph please
Answer:
1. W = F d = 20 N * 6 m = 120 J
2. F = W / d = 60 J / 2 m = 30 N
3. d = W / F = 350 J / 85 N = 4.12 m
4. P = W / t = F d / t = 45 N * 9 m / 10 s = 40.5 Watts
5. W = P t = 500 W * 120 sec = 60,000 J
6. t = W / P = 550 J / 310 W = 1.77 sec
Answer:
2.464 cm above the water surface
Explanation:
Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.
We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3 cm^3):
weight of the block = 0.78 * 11.2^3 gr
Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.
So the weight of the volume of water displaced is:
weight of water = 1 * 11.2^2 * x
we make both weight expressions equal each other for the floating requirement:
0.78 * 11.2^3 = 11.2^2 * x
then x = 0.78 * 11.2 cm = 8.736 cm
This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:
11.2 cm - 8.736 cm = 2.464 cm
