Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
Answer:
option 1
Explanation:
i just used the SOH CAH TOA, and since the given is tan=opposite/adjacent, that should be the answer
Answer:
Yes, yes it would since we need light
Explanation:
Answer:opposite
Explanation:for a capacitor to discharge (after charging) the polarities of the current and voltage have to be reversed
Answer:
m = 0.51[kg]
Explanation:
Potential energy is defined as the product of mass by gravity by height.
where:
Epot = potential energy = 15 [J]
m = mass [kg]
g = gravity acceleration = 9.8 [m/s²]
h = elevation = 3 [m]
Now replacing:
![E_{pot}=m*g*h\\15=m*9.8*3\\m = 0.51[kg]](https://tex.z-dn.net/?f=E_%7Bpot%7D%3Dm%2Ag%2Ah%5C%5C15%3Dm%2A9.8%2A3%5C%5Cm%20%3D%200.51%5Bkg%5D)
