Answer:
The average forces would be the same
Explanation:
Both have the same velocity on impact as they fell from the same height.
Both have the same velocity after the bounce because they reach the same height.
Both have the same mass
Both will thus experience the same impulse because both have the same change in momentum.
Therefore both experience the same average force.
Answer:v=41.23 m/s
Explanation:
Given
mass of heavy object ![m_1=52 kg](https://tex.z-dn.net/?f=m_1%3D52%20kg)
distance of
from the axle ![r_1=14 cm](https://tex.z-dn.net/?f=r_1%3D14%20cm)
mass of rock ![m_2=123 gm](https://tex.z-dn.net/?f=m_2%3D123%20gm)
Length of rod ![=4.1 m](https://tex.z-dn.net/?f=%3D4.1%20m%20)
distance of
from axle ![r_2=4.1-0.14=3.96 m](https://tex.z-dn.net/?f=r_2%3D4.1-0.14%3D3.96%20m)
Net torque acting is
![T_{net}=m_1gr_1-m_2gr_2](https://tex.z-dn.net/?f=T_%7Bnet%7D%3Dm_1gr_1-m_2gr_2)
![T_{net}=52\times 0.14\times g-0.123\times 3.96\times g](https://tex.z-dn.net/?f=T_%7Bnet%7D%3D52%5Ctimes%200.14%5Ctimes%20g-0.123%5Ctimes%203.96%5Ctimes%20g)
![T_{net}=6.793\times 9.8](https://tex.z-dn.net/?f=T_%7Bnet%7D%3D6.793%5Ctimes%209.8)
![T_{net}=66.57 N-m](https://tex.z-dn.net/?f=T_%7Bnet%7D%3D66.57%20N-m)
Work done by
is converted to rock kinetic Energy
thus
![T_{net}\times \theta =\frac{mv^2}{2}](https://tex.z-dn.net/?f=T_%7Bnet%7D%5Ctimes%20%5Ctheta%20%3D%5Cfrac%7Bmv%5E2%7D%7B2%7D)
Where ![\theta =angle\ turned =\frac{\pi }{2}](https://tex.z-dn.net/?f=%5Ctheta%20%3Dangle%5C%20turned%20%3D%5Cfrac%7B%5Cpi%20%7D%7B2%7D)
![v= velocity\ at\ launch](https://tex.z-dn.net/?f=v%3D%20velocity%5C%20at%5C%20launch)
![66.57\times \frac{\pi }{2}=\frac{0.123\times v^2}{2}](https://tex.z-dn.net/?f=66.57%5Ctimes%20%5Cfrac%7B%5Cpi%20%7D%7B2%7D%3D%5Cfrac%7B0.123%5Ctimes%20v%5E2%7D%7B2%7D)
![v^2=66.57\times \pi](https://tex.z-dn.net/?f=v%5E2%3D66.57%5Ctimes%20%5Cpi%20)
![v=\sqrt{1700.511}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B1700.511%7D)
![v=41.23 m/s](https://tex.z-dn.net/?f=v%3D41.23%20m%2Fs)
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
Answer:
1) D
2) A
Explanation:
1) Each group has the same number of valence electrons, which are the outer electrons.
2) Ionic bonds are between a metal and non - metal, the metal being sodium and the non - metal being chlorine.
The answer is 21m because the motion is in one dimension with constant acceleration.
The initial velocity is 0, because it started from rest, the acceleration <span>ax</span> is <span>4.7<span>m<span>s2</span></span></span>, and the time t is <span>3.0s</span>
Plugging in our known values, we have
<span>Δx=<span>(0)</span><span>(3.0s)</span>+<span>12</span><span>(4.7<span>m<span>s2</span></span>)</span><span><span>(3.0s)</span>2</span>=<span>21<span>m</span></span></span>