Answer:
a) 10.7° ≈ 11°
b) 0.19
Explanation:
If the road is banked at an angle, without seeking the help of friction, (i.e. frictionless road), the forces acting on the car are shown in the attached free body diagram to the question
In the y - direction
mg = N cos θ (eqn 1)
mg = weight of the car.
N = normal reaction of the plane on the car
And in the direction parallel to the inclined plane,
(mv²/r) = N sin θ (eqn 2)
(mv²/r) = force keeping the car in circular motion
Divide (eqn 2) by (eqn 1)
(v²/gr) = Tan θ
v = velocity of car = 60 km/h = 16.667 m/s
g = acceleration due to gravity
r = 150 m
(16.667²/(9.8×150)) = Tan θ
θ = Tan⁻¹ (0.18896)
θ = 10.7° ≈ 11°
b) In the absence of banking, the frictional force on the road has to balance the force keeping the car in circular motion
That is,
Fr = (mv²/r)
Fr = μN = μ mg
μ mg = mv²/r
μ = (v²/gr) = (16.667²/(9.8×150)) = 0.19
Hope this Helps!!!
It will take 21s for yhe alligator to crawl that distance. Reported answer contains 2 Significant Figures.
Answer:
II) Kitchen waste: Meal leftovers, Banna peelings
Garden Waste: Camote leaves, Kangkong leaves, weeds
Factory: Glass bottles, carton pieces
III) A
IV) Home: Bottles of shampoo, leftover food, syringe
office Gloves
Classroom: containers
Laboratory: empty cartridge
Answer:
Explanation:
To find the angular velocity of the tank at which the bottom of the tank is exposed
From the information given:
At rest, the initial volume of the tank is:
where;
height h which is the height for the free surface in a rotating tank is expressed as:
at the bottom surface of the tank;
r = 0, h = 0
∴
0 = 0 + C
C = 0
Thus; the free surface height in a rotating tank is:
Now; the volume of the water when the tank is rotating is:
dV = 2π × r × h × dr
Taking the integral on both sides;
replacing the value of h in equation (2); we have:
![V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0](https://tex.z-dn.net/?f=V_f%20%3D%20%5Cdfrac%7B%20%5Cpi%20%5Comega%20%5E2%7D%7Bg%7D%20%5CBig%20%5B%20%20%5Cdfrac%7Br%5E4%7D%7B4%7D%20%5CBig%5D%5ER_0)
![V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)](https://tex.z-dn.net/?f=V_f%20%3D%20%5Cdfrac%7B%20%5Cpi%20%5Comega%20%5E2%7D%7Bg%7D%20%5CBig%20%5B%20%20%5Cdfrac%7BR%5E4%7D%7B4%7D%20%5CBig%5D%20---%20%283%29)
Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.
Then 
Replacing equation (1) and (3)
Finally, the angular velocity of the tank at which the bottom of the tank is exposed = 10.48 rad/s
Answer:
10.2 kg
Explanation:
Assuming the object is on Earth:
w = mg
m = w / g = 100 / 9.81 = 10.2kg
