Answer:
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=
r
1
mv
2
=
r
2
mu
2
................ii
form i and ii we can write
v^2= \frac{1}{2} u^2v
2
=
2
1
u
2
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
Answer:
x^3(15x^4y^2+4)
Step-by-step explanation:
Answer:
27.85 remainder 7
Step-by-step explanation: devide something
5. Given the equation y = (1/4) cos[(2pi/3)*theta]:
5a. For the general equation y = a cos(bx), the period is given by 2pi/b. In this equation, b = 2pi/3, so this means that 2pi/b = 2pi/(2pi/3) = 3. Therefore, the period of this equation is 3, and the cosine wave repeats itself every 3 x-units.
5b. For the general equation y = a cos(bx), the amplitude is given by a. Therefore the amplitude is a = 1/4, and this means that the cosine wave's range is from -1/4 to 1/4 for all values of x.
5c. The equation of the midline is y = 0. This represents the average value over the wave. This is determined by adding the highest and lowest values of the range and taking the average, in this case, 1/4 + (-1/4) = 0, and 0 / 2 = 0. Another way to do this is using the general equation y = a cos(bx) + c, where the midline's equation is y = c. In this case, there is no value of c in the given, implying that c = 0, and the midline is y = 0.
6. Let the horizontal distance be x. Then tan42 = h/x, and h = x tan42. Then using the Pythagorean theorem: 3280^2 = h^2 + x^2
3280^2 = x^2 (tan42)^2 + x^2
3280^2 = x^2 [(tan42)^2 + 1]
x = 2437.52
Therefore, h = x tan42 = 2,194.75 ft.
Answer:
0.15
Step-by-step explanation:
9.70-8.95= 0.75
0.75÷5= 0.15
