Answer:
a) a = 3.53 m/sec²
b ) As force in the sense of movement is bigger than Fr (that is m*g*sin21 > Fr ) the car will slide down.
c) β = 11⁰
Step-by-step explanation: We must start calculating the angle of the road.
37% means that inclination is such that for 100 m of horizontal distance we will be up the path a 37 m, then tan α = 0.37 and α = 21⁰
a) According to Newtons second law
∑ Fx = ma ⇒ if only gravity is acting
m*g* sinα = ma ⇒ 1000 Kg*9.8 m/sec² *sin 21° = 1000 Kg*a
9.8 * 0,36 = a ⇒ a = 3.53 m/sec²
b) Fr = μ*Fn
∑ Fy = 0
m*g*cos21° - Fn = 0 Fn = 1000*9.8*cos21° Fn = 9143.4 [N]
Then Fr = 9143.4 * 0,3 ⇒ Fr = 2743.02 [N]
And
m*g*sinα = m*g*sin21⁰ = 1000* 9.8* 0.36 = 3528 [N]
As force in the sense of movement is bigger than Fr (that is m*g*sin21 > Fr ) the car will road down.
c) If μk = 0.2
m*g*sinβ = 9800* sinβ - 9143.4*0.2 = 0
sinβ = 1828.68/9800 ⇒ sinβ = 0.1866 ⇒ β = 11⁰
The answer should be (-1/3,3/4) if I did the math correct
Answer: The length is 15 yards so the answer is A.
Step-by-step explanation:
If the length is 3 more than the width the we will represent it by the equation L = w+3 where w is the width and to find the area of a rectangle, we need to multiply the length by the width. so the length is w+3 and the width is w so
w(w+3) = 180 solve for the w
w^2 + 3w = 180 subtract 180 from both sides
w^2 + 3w -180 = 0 find two numbers that their product is -180 and add to 3.
the number 12 and -15 works out because -12*15= -180 and -15+ 12 =3
We will now have a new quadratic equation as
w^2 - 12w + 15w - 180 = 0 factor by grouping
w(w-12) 15(w-12)= 0 factor out w-12
(w-12)(w+15) = 0 set them both equal zero.
w -12 = 0 or w+15= 0
w = 12 or w = -15
Since we know that -15 can't represent a distance, then 12 is the width .
So if we are to find the length and it gives us the information that the length is 3 yards more that the width then we will add 3 to the width to equal the length.
L= 12 +3
L = 15
Is this an actual question or a joke? if its a joke idk what?
No. All segments have one perpendicular bisector.