Air resistance, also called drag, acts upon a falling body by slowing the body down to thr point where it stops accelerating, and it falls at a constant speed, known as the terminal volocity of a falling object. Air resistance depends on the cross sectional area of the object, which is why the effect of air resistance on a large flat surfaced object is much greater than on a small, streamlined object.
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>
Wish I could help you you rock
The gravitational force is inversely proportional to the
square of the distance between their centers. So the
force is greatest when the distance is zero.
Answer:
R = 98304.75 m = 98.3 km
Explanation:
The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.
Density = Mass/Volume
Now, it is given that the density of Earth has become:
Density = 1 x 10⁹ kg/m³
Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg
Volume = 4/3πR³ (Volume of Sphere)
R = Radius of Earth = ?
Therefore,
1 x 10⁹ kg/m³ = (5.97 x 10²⁴ kg)/[4/3πR³]
4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)
R³ = (3/4)(5.97 x 10¹⁵ m³)/π
R = ∛[0.95 x 10¹⁵ m³]
<u>R = 98304.75 m = 98.3 km</u>
