Answer:
because it’s suppose to be red like a stop light.
Explanation:
So it tells you to stop
Answer:
7.1 m/s
Explanation:
First, find the time it takes for the fish to reach the water.
Given in the y direction:
Δy = 6.1 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
6.1 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 1.12 s
Next, find the velocity needed to travel 7.9 m in that time.
Given in the x direction:
Δx = 7.9 m
a = 0 m/s²
t = 1.12 s
Find: v₀
Δx = v₀ t + ½ at²
7.9 m = v₀ (1.12 s) + ½ (0 m/s²) (1.12 s)²
v₀ = 7.1 m/s
Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s
Answer:
Direction of ship: 9.45° West of North
Ship's relative speed: 7.87m/s
Explanation:
A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0
Vx=0
Therefore, -VsSin∅+VcCos∅40°
Sin∅ = Vc/Vs × Cos 40°
Sin∅ = 1.5/7 ×Cos40°
Sin∅= 0.164
∅= Sin-¹ (0.164)
∅= 9.45° W of N
B. Ship's relative speed:
Vy= VsCos∅ + Vcsin40°
= 7Cos9.45° + 1.5sin40°
= 7×0.986 + 1.5×0.642
= 7.865
= 7.87m/s
The right answer is "Strong nuclear force"