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>
W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s
Answer:
1.97 * 10^8 m/s
Explanation:
Given that:
n = 1.52
Recall : speed of light (c) = 3 * 10^8 m/s
Speed (v) of light in glass:
v = speed of light / n
v = (3 * 10^8) / 1.52
v = 1.9736 * 10^8
Hence, speed of light in glass :
v = 1.97 * 10^8 m/s
Answer:
Here ball and rod will repel each other as they are of similar charges
Explanation:
As we know that the two charges attract or repel each other by electrostatic force
This force is given as
so we know if two charges are similar in nature then they will repel each other and if the two charges are opposite in nature then they will attract each other
So here when rod touch the ball then it transfer its charge to the ball and due to similar charges in ball and rod they both repel each other
