Answer:
Tires.
Explanation:
There are the few steps which are discussed below should be taken to increase or extend the life of tires.
(1) Avoid fast starts: Fast start of the vehicle will increase the pressure on the tires due to the friction between the tires and the road will decrease the life of tires.
(2) Avoid fast stop: Fast stop of the vehicle will also increase the pressure on the tires due to the friction between the tires and the road will decrease the life of tires.
(3) Avoid sharp turns: The alignment of the wheels and tires are in such a way that they work properly when vehicle is drive in a straight path but sharp turn will increase the uneven pressure on the tires will lead to decrease the life of tires.
Therefore, the life of tires can be extend by avoiding all the above mention actions such as fast stop, start and sharp turns.
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:
s = 23.72 m
v = 21.56 m/s²
Explanation:
given
time to reach the ground (t) = 2.2 second
we know that
a) s = u t + 0.5 g t²
u = 0 m/s
g = 9.8 m/s²
s = 0 + 0.5 × 9.8 × 2.2²
s = 23.72 m
b) impact velocity
v = √(2gh)
v = √(2× 9.8 × 23.72)
v = √464.912
v = 21.56 m/s²
This is called the Phi Phenomenon.
This is an illusion of movement created when two or more adjacent lights blink on and off in quick succession; when two adjacent stationary lights blink on and off in quick succession; we perceive a single light moving back and forth between them. It is an optical illusion of perceiving a series of still images, when viewed in rapid succession, as continuous motion.
