If the coefficient of friction between road and tires on a rainy day is 0.109, what is the minimum distance in which the car wil
1 answer:
The car will halt at a minimum distance of 98.57 meters on a rainy day having a friction coefficient of 0.109.
Calculation of the minimum distance-
Provided that :
- the speed of the car = 52 km/h
= 52 x 0.278 m/s = 14.45 m/s
- Friction coefficient, μ = 0.109
the regular force exerted on the car,
Along X-direction, the force is
Friction acts in the opposite way along the x-axis
⇒-μ
⇒-μmg =
⇒ μg
Utilizing the motion equation-
v² = u²+ 2 .a. s
The final speed, v=0 m/s
⇒0² = (14.45)² - 2 μg .s
⇒2 * 0.109 *9.8 *s = (14.45)² = 208.8
⇒s = 208.8 / 2.14 = 97.57 m
It is concluded that the car will halt at a minimum distance of 98.57 meters.
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Answer:
a) V = 0.085 m^3
b) m = 17 kg
Explanation:
1) Data given
mb = 68 kg (mass for the block)
20% of the block volume is floating
100-20= 80% of the block volume is submerged
2) Notation
mb= mass of the block
Vw= volume submerged
mw = mass water displaced
V= total volume for the block
3) Forces involved (part a)
For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)
Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :
B = (mass water displaced) g = (mw) g (1)
The definition of density is :
If we solve for mw we got (2)
Replacing equation (2) into equation (1) we got:
(3)
On this case Vw represent the volume of water displaced = 0.8 V
If we replace the values into equation (3) we have
0.8 ρ_w V g = mg (4)
And solving for V we have
V = (mg)/(0.8 ρ_w g )
We cancel the g in the numerator and the denominator we got
V = (m)/(0.8 ρ_w)
V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3
4) Forces involved (part b)
For this case we have bricks above the block, and we want the maximum mass for the bricks without causing it to sink below the water surface.
We can begin finding the weight of the water displaced when the block is just about to sink (W1)
W1 = ρ_w V g
W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N
After this we can calculate the weight of water displaced before putting the bricks above (W2)
W2 = 0.8 x 833 N = 666.4 N
So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)
W3 = W1 -W2 = 833-666.4 N = 166.6 N
And finding the mass fro the definition of weight we have
m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg