Answer:
you can predict where the juggling ball is going to land and the move you hand to catch it
Explanation:
Answer:
![\boxed {\tt 3.63636364 \ m/s}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Ctt%203.63636364%20%5C%20m%2Fs%7D)
Explanation:
Velocity can be found using the following formula:
![v=\frac{p}{m}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bp%7D%7Bm%7D)
where p is the momentum and m is the mass.
The woman has a mass of 55 kilograms and a momentum of 200 kilogram meters per second.
![p= 200 \ kgm/s\\m=55 \ kg](https://tex.z-dn.net/?f=p%3D%20200%20%5C%20kgm%2Fs%5C%5Cm%3D55%20%5C%20kg)
Substitute the values into the formula.
![v=\frac{200 \ kg m/s}{55 \ kg}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B200%20%5C%20kg%20m%2Fs%7D%7B55%20%5C%20kg%7D)
Divide. Note that the kilograms, or kg, will cancel each other out.
![v=\frac{200 \ m/s}{55}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B200%20%5C%20m%2Fs%7D%7B55%7D)
![v= 3.63636364 \ m/s](https://tex.z-dn.net/?f=v%3D%203.63636364%20%5C%20m%2Fs)
The woman's velocity is 3.63636364 meters per second.
Answer: Line graph should be used to show how one variable changes over time not to show multiple categories or variables are at one specific point in time.
Explanation:
In maths, statistics, and related fields, graphs are used to visually display variables and their values. In the case of line graphs, these are mainly used to display evolution or change of a variable over time. For example, a line graph can show how the number of divorces changed from 1920 to 2010.
In this context, the number of different animals in the park cannot be represented through a line graph because this situation does not imply a variable changing over time. Moreover, this situation includes multiple variables or categories of animals and the data shows only one specific point in time, which can be better represented through a bar graph.
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