A bond with elements from B.
Mass of the object is given as
![m = 1417 g = 1.417 kg](https://tex.z-dn.net/?f=m%20%3D%201417%20g%20%3D%201.417%20kg)
now the speed of object is given as
![v = \frac{d}{t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bd%7D%7Bt%7D)
here we know that
![d = 47 m](https://tex.z-dn.net/?f=d%20%3D%2047%20m)
![t = 90 s](https://tex.z-dn.net/?f=t%20%3D%2090%20s)
now we will have
![v = \frac{47}{90} = 0.52 m/s](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B47%7D%7B90%7D%20%3D%200.52%20m%2Fs)
now we will have kinetic energy of the object as
![KE = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![KE = \frac{1}{2}(1.417)(0.52)^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%281.417%29%280.52%29%5E2)
![KE = 0.19 J](https://tex.z-dn.net/?f=KE%20%3D%200.19%20J)
now the power is defined as rate of energy
so here we can find power as
![P = \frac{KE}{t}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BKE%7D%7Bt%7D)
![P = \frac{0.19}{90} = 2.14\times 10^{-3} W](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B0.19%7D%7B90%7D%20%3D%202.14%5Ctimes%2010%5E%7B-3%7D%20W)
so above is the power used for the object
You will use the Pythagorean Theorem to solve it.
c^2 = a^2 + b^2
c^2 = (1.5)^2 + (2)^2
c^2 = 6.25
c = square root of 6.25
c = 2.5
I hope this helps!
Answer:
The x represents the reference point on a motion map
Explanation:
-Motion maps are another way to represent the motion of an object. (other representations are graphical and mathematical models)
Answer:
48.16 %
Explanation:
coefficient of restitution = 0.72
let the incoming speed be = u
let the outgoing speed be = v
kinetic energy = 0.5 x mass x ![x velocity^{2}](https://tex.z-dn.net/?f=%20x%20velocity%5E%7B2%7D)
- incoming kinetic energy = 0.5 x m x
- coefficient of restitution =
![\frac{v}{u}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%7D%7Bu%7D)
0.72 =![\frac{v}{u}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%7D%7Bu%7D)
v = 0.72u
therefore the outgoing kinetic energy = 0.5 x m x ![(0.72u)^{2}](https://tex.z-dn.net/?f=%280.72u%29%5E%7B2%7D)
outgoing kinetic energy = 0.5 x m x ![0.5184 x u^{2}](https://tex.z-dn.net/?f=0.5184%20x%20u%5E%7B2%7D)
outgoing kinetic energy = 0.5184 (0.5 x m x
)
recall that 0.5 x m x
is our incoming kinetic energy, therefore
outgoing kinetic energy = 0.5184 x (incoming kinetic energy)
from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.
The energy lost would be 100 - 51.84 = 48.16 %