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:
Isotope it will have a different number of neutrons than normal
Explanation:
Given Data
Total mass=93.5 kg
Rock mass=0.310 kg
Initially wagon speed=0.540 m/s
rock speed=16.5 m/s
To Find
The speed of the wagon
Solution
As the wagon rolls, momentum is given as
P=mv
where
m is mass
v is speed
put the values
P=93.5kg × 0.540 m/s
P =50.49 kg×m/s
Now we have to find the momentum of rock
momentum of rock = mv
momentum of rock = (0.310kg)×(16.5 m/s)
momentum of rock =5.115 kg×m/s
From the conservation of momentum we can find the wagons momentum So
wagon momentum=50.49 -5.115 = 45.375 kg×m/s
Speed of wagon = wagon momentum/(total mass-rock mass)
Speed of wagon=45.375/(93.5-0.310)
Speed of wagon= 0.487 m/s
Throwing rock backward,
momentum of wagon = 50.49+5.115 = 55.605 kg×m/s
Speed of wagon = wagon momentum/(total mass-rock mass)
speed of wagon = 55.605 kg×m/s/(93.5kg-0.310kg)
speed of wagon= 0.5967 m/s
Answer:
27 m/s
Explanation:
Given:
v₀ = 15 m/s
a = 3 m/s²
t = 4 s
Find: v
v = at + v₀
v = (3 m/s²) (4 s) + (15 m/s)
v = 27 m/s
Answer:
A mass of 4 Kg rest on the horizontal plane. The plane is gradually inclined until at an angle of θ=15
∘
with the horizontal,the mass just being to slide what is the coffeficient of static friction between the block & the surface.
Explanation:
