the Orbital Velocity is the velocity sufficient to cause a natural or artificial satellite to remain in orbit. Inertia of the moving body tends to make it move on in a straight line, while gravitational force tends to pull it down. The orbital path, elliptical or circular, representing a balance between gravity and inertia, and it follows a rue that states that the more massive the body at the centre of attraction is, the higher is the orbital velocity for a particular altitude or distance.
The answer is attract. Hope it helps! :)
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
Answer: Light passes through the front of the eye (cornea) to the lens. The cornea and the lens help to focus the light rays onto the back of the eye (retina). The cells in the retina absorb and convert the light to electrochemical impulses which are transferred along the optic nerve and then to the brain.
Explanation:
When light rays reflect off an object and enter the eyes through the cornea (the transparent outer covering of the eye), you can then see that object. The cornea bends, or refracts, the rays that pass through the round hole of the pupil.
Answer:
Explanation:
We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:
The acceleration can be calculated using Newton's second law:
Recall that the maximum force of friction is defined as 
. So, replacing this:
Now, we calculate the distance:
