Technically friction is acting on the car because it is still rubbing against the street and gravity is pulling the car down preventing it from floating??? lol
Answer:
E means energy
M= Mass
C=speed of light squared (the exponent means squared)
Answer:
velocity = 1527.52 ft/s
Acceleration = 80.13 ft/s²
Explanation:
We are given;
Radius of rotation; r = 32,700 ft
Radial acceleration; a_r = r¨ = 85 ft/s²
Angular velocity; ω = θ˙˙ = 0.019 rad/s
Also, angle θ reaches 66°
So, velocity of the rocket for the given position will be;
v = rθ˙˙/cos θ
so, v = 32700 × 0.019/ cos 66
v = 1527.52 ft/s
Acceleration is given by the formula ;
a = a_r/sinθ
For the given position,
a_r = r¨ - r(θ˙˙)²
Thus,
a = (r¨ - r(θ˙˙)²)/sinθ
Plugging in the relevant values, we obtain;
a = (85 - 32700(0.019)²)/sin66
a = (85 - 11.8047)/0.9135
a = 80.13 ft/s²
Answer:
a) x(t) = 10t + (2/3)*t^3
b) x*(0.1875) = 10.18 m
Explanation:
Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.
Given:
- v(t) = 10 + 2*t^2 (radar gun)
- x*(t) = 10 + 5t^2 + 3t^3 (our coordinate)
Find:
-The position x of horse as a function of time t in radar system.
-The position of the horse at x = 2m in our coordinate system
Solution:
- The position of horse according to radar gun:
v(t) = dx / dt = 10 + 2*t^2
- Separate variables:
dx = (10 + 2*t^2).dt
- Integrate over interval x = 0 @ t= 0
x(t) = 10t + (2/3)*t^3
- time @ x = 2 :
2 = 10t + (2/3)*t^3
0 = 10t + (2/3)*t^3 + 2
- solve for t:
t = 0.1875 s
- Evaluate x* at t = 0.1875 s
x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3
x*(0.1875) = 10.18 m
