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: Force = Mass X Acceleration
F = 5 x 2
F = 10 N
Answer:
gather the data needed to make the forecast
Fahrenheit because the boiling point of water is 100 degrees Celsius which is 212 Fahrenheit which is very hot, and that would be about 200 Kelvin so therefore the answer is that the temperature was recorded in Fahrenheit not Kelvin or Celsius
Answer:
a)Q=71.4 μ C
b)ΔV' = 10.2 V
Explanation:
Given that
C ₁= 8.7 μF
C₂ = 8.2 μF
C₃ = 4.1 μF
The potential difference of the battery, ΔV= 34 V
When connected in series
1/C = 1/C ₁ + 1/C₂ + 1/C₃
1/ C= 1/8.4 +1 / 8.4 + 1/4.2
C=2.1 μF
As we know that when capacitor are connected in series then they have same charge,Q
Q= C ΔV
Q= 2.1 x 34 μ C
Q=71.4 μ C
b)
As we know that when capacitor are connected in parallel then they have same voltage difference.
Q'= C' ΔV'
C'= C ₁+C₂+C₃ (For parallel connection)
C'= 8.4 + 8.4 + 4.2 μF
C'=21 μF
Q'= C' ΔV'
Q'=3 Q
3 x 71.4= 21 ΔV'
ΔV' = 10.2 V
