Let u = the speed of the car at the instant when braking begins.
The braking distance is s = 62.3 m, the acceleration is a = -5.9 m/s², and the braking duration is t = 4.15 s.
Use the formula s = ut + (1/2)at² to obtain
(u m/s)*(4.15 s) + 0.5*(-5.9 m/s²)*(4.5 s)² = (62.3 m)
4.15u = 62.3 + 50.8064 = 113.1064
u = 27.2546 m/s
Let v m/s be the speed with which the car strikes the tree.
Then
v = 27.2546 - 5.9*4.15
= 2.7696 m/s
Answer: 2.77 m/s (nearest hundredth)
Fun fhjzsh going chichi. Gok
Answer:
V₀ = 5.47 m/s
Explanation:
The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:
R = V₀² Sin 2θ/g
where,
R = Range of Projectile = 3.04 m
θ = Launch Angle = 41.7°
V₀ = Minimum Launch Speed = ?
g = 9.81 m/s²
Therefore,
3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)
V₀² = 3.04 m/(0.10126 s²/m)
V₀ = √30.02 m²/s²
<u>V₀ = 5.47 m/s</u>
Answer:
okay
Explanation:
(wish i could help but im just answering for points)
Answer:
balanced?
Explanation:
because if it wasn't moving that means they are pulling at a similar strength
