Answer:
2.45 m
Explanation:
We are given that
![m_1=130 kg](https://tex.z-dn.net/?f=m_1%3D130%20kg)
![m_2=72 kg](https://tex.z-dn.net/?f=m_2%3D72%20kg)
Height,h=0.75 m
Initial velocity,![u_1=u_2=0](https://tex.z-dn.net/?f=u_1%3Du_2%3D0)
We have to find the height above his own starting point Ed rises.
Initial kinetic energy of Ed=Final potential energy of Ed
![m_2gh'=\frac{1}{2}m_2v^2_2](https://tex.z-dn.net/?f=m_2gh%27%3D%5Cfrac%7B1%7D%7B2%7Dm_2v%5E2_2)
![h'=\frac{v^2_2}{2g}](https://tex.z-dn.net/?f=h%27%3D%5Cfrac%7Bv%5E2_2%7D%7B2g%7D)
According to law of conservation of momentum
![0=m_1v_1+m_2v_2](https://tex.z-dn.net/?f=0%3Dm_1v_1%2Bm_2v_2)
![v_2=-\frac{m_1v_1}{m_2}](https://tex.z-dn.net/?f=v_2%3D-%5Cfrac%7Bm_1v_1%7D%7Bm_2%7D)
![h'=\frac{m^2_1v^2_1}{2gm^2_2}](https://tex.z-dn.net/?f=h%27%3D%5Cfrac%7Bm%5E2_1v%5E2_1%7D%7B2gm%5E2_2%7D)
Initial kinetic energy of adolf=Final potential energy of adolf
![\frac{1}{2}m_1v^2_1=m_1gh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dm_1v%5E2_1%3Dm_1gh)
![v_1=\sqrt{2gh}](https://tex.z-dn.net/?f=v_1%3D%5Csqrt%7B2gh%7D)
Substitute the values
![h'=\frac{(130)^2(\sqrt{2\times 9.8\times 0.75})^2}{(72)^2\times 2\times 9.8}](https://tex.z-dn.net/?f=h%27%3D%5Cfrac%7B%28130%29%5E2%28%5Csqrt%7B2%5Ctimes%209.8%5Ctimes%200.75%7D%29%5E2%7D%7B%2872%29%5E2%5Ctimes%202%5Ctimes%209.8%7D)
![h'=2.45 m](https://tex.z-dn.net/?f=h%27%3D2.45%20m)
To find the relative frequencies, divide each frequency by the total number of throws - in this case, 100.
Red: 10/100=1/10=0.1
Yellow:35/100=0.35
Blue:48/100= 0.48
Misses: 7/100=0.07
The event of landing in the red region has a relative frequency of 0.1 which means the dart landed in the red region about 10% (0.1 x 100%) of the time.
Answer:
0.947 rad or 54.27 degrees
Explanation:
Suppose the collision is elastic, meaning the momentum is preserved.
Before the collision
and the second ball has no momentum
Right after the collision
and the first bast ball has no momentum.
Therefore momentum of the first baseball has been transferred to the 2nd ball
![m_1v_1 = m_2v_2](https://tex.z-dn.net/?f=%20m_1v_1%20%3D%20m_2v_2)
![4.5 * 0.45 = 0.61v_2](https://tex.z-dn.net/?f=%204.5%20%2A%200.45%20%3D%200.61v_2)
![v_2 = \frac{4.5 * 0.45}{0.61} = 3.32 m/s](https://tex.z-dn.net/?f=%20v_2%20%3D%20%5Cfrac%7B4.5%20%2A%200.45%7D%7B0.61%7D%20%3D%203.32%20m%2Fs)
After the collision, the second ball would gained kinetic energy, which then would be transferred to potential energy once it reaches its highest point:
By the law of energy conservation:
By law of energy conservation:
![y = \frac{v^2}{2g} = \frac{3.32^2}{2*9.81} = 0.562 m](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7Bv%5E2%7D%7B2g%7D%20%3D%20%5Cfrac%7B3.32%5E2%7D%7B2%2A9.81%7D%20%3D%200.562%20m)
So at its highest point, the ball is 0.562 m from the lowest point. Since the ball is hanging on a 1.35 m string, we can calculate the vertical distance from there to the swinging point:
1.35 - 0.562 = 0.788 m
Finally, the angle that string makes with the vertical at the highest point is
![\alpha = cos^{-1}(\frac{0.788}{1.35}) = 0.947 rad = 54.27^o](https://tex.z-dn.net/?f=%5Calpha%20%3D%20cos%5E%7B-1%7D%28%5Cfrac%7B0.788%7D%7B1.35%7D%29%20%3D%200.947%20rad%20%3D%2054.27%5Eo)
Before we solve, let's use our formula for speed.
Speed is defined as distance divided by time.
S = D/T
We have 119m and 28s.
Our distance is 119m.
Our time is 28s.
Plug these numbers into the formula.
S = 119/28
Divide 119 by 28 to get our speed.
119/28 = 4.25.
Our speed is 4.25 m/s.
I hope this helps!