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)
Answer:
The helicopter uses 35 gallons to fly for 5 hours.
Explanation:
The amount of gas that a helicopter uses for flying varies directly proportional to the number of hours spent flying.
g ∝ T
where g represents amount of gas and T time of flight.
Then,

The helicopter files 4 hours and uses 28 gallons of fuel.
Here, g₁= 28 gallons, T₁=4 hours
g₂=?, T₂=5 hours.


⇒28×5= g₂×4
⇒ g₂×4=28×5

gallons
The helicopter uses 35 gallons to fly for 5 hours.
If im not mistaking its the last one slowing heat transfer from the inside to the outside of the container
Answer:
the pressure at B is 527psf
Explanation:
Angular velocity, ω = v / r
ω = 20 /1.5
= 13.333 rad/s
Flow equation from point A to B
![P_A+rz_A-\frac{1}{2} Pr_A^2w^2=P_B+rz_B-\frac{1}{2} pr^2_Bw^2\\\\P_B = P_A + r(z_A-z_B)+\frac{1}{2} pw^2[(r_B^2)-(r_A)^2]\\\\P_B = [25 +(0.8+62.4)(0-1)+\frac{1}{2}(0.8\times1.94)\times(13.333)^2[2.5^2-1.5^2] ]\\\\P_B = 25 - 49.92+551.79\\\\P_B = 526.87psf\\\approx527psf](https://tex.z-dn.net/?f=P_A%2Brz_A-%5Cfrac%7B1%7D%7B2%7D%20Pr_A%5E2w%5E2%3DP_B%2Brz_B-%5Cfrac%7B1%7D%7B2%7D%20pr%5E2_Bw%5E2%5C%5C%5C%5CP_B%20%3D%20P_A%20%2B%20r%28z_A-z_B%29%2B%5Cfrac%7B1%7D%7B2%7D%20pw%5E2%5B%28r_B%5E2%29-%28r_A%29%5E2%5D%5C%5C%5C%5CP_B%20%3D%20%5B25%20%2B%280.8%2B62.4%29%280-1%29%2B%5Cfrac%7B1%7D%7B2%7D%280.8%5Ctimes1.94%29%5Ctimes%2813.333%29%5E2%5B2.5%5E2-1.5%5E2%5D%20%20%5D%5C%5C%5C%5CP_B%20%3D%2025%20-%2049.92%2B551.79%5C%5C%5C%5CP_B%20%3D%20526.87psf%5C%5C%5Capprox527psf)
the pressure at B is 527psf