I think that the shopping cart full of groceries has more inertia because it is the one with more tendency to do nothing or be still.
Answer: a) 19.21m b) 3.92secs
Explanation:
a) Maximum height reached by the object is the height reached by an object before falling freely under gravity.
Maximum height = U²/2g
U is the initial velocity = 19.6m/s
g is acceleration due to gravity = 10m/s²
Maximum Height = 19.6²/2(10)
H = 19.21m
b) The time elapsed before the stone hits the ground is the time of flight T= 2U/g
T= 2(19.6)/10
T = 39.2/10
Time elapsed is 3.92secs
Image #3 good luck!!!!!!!!!!!
Answer:
The copper atoms are heavier than the aluminum atoms. The copper atoms are smaller than the aluminum atoms so more copper atoms fit in the same volume. 2. Copper is more dense than aluminum.
HOPE THIS HELPS !!!!!
Answer:
![h = 18.75 m](https://tex.z-dn.net/?f=h%20%3D%2018.75%20m)
Now when it will reach at point B then its normal force is just equal to ZERO
![N_B = 0](https://tex.z-dn.net/?f=N_B%20%3D%200)
![F_n = 1.72 \times 10^4](https://tex.z-dn.net/?f=F_n%20%3D%201.72%20%5Ctimes%2010%5E4)
Explanation:
Since we need to cross both the loops so least speed at the bottom must be
![v = \sqrt{5 R g}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B5%20R%20g%7D)
also by energy conservation this is gained by initial potential energy
![mgh = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![v = \sqrt{2gh}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2gh%7D)
so we will have
![\sqrt{2gh} = \sqrt{5Rg}](https://tex.z-dn.net/?f=%5Csqrt%7B2gh%7D%20%3D%20%5Csqrt%7B5Rg%7D)
now we have
![h = \frac{5R}{2}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B5R%7D%7B2%7D)
here we have
R = 7.5 m
so we have
![h = \frac{5(7.5)}{2}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B5%287.5%29%7D%7B2%7D)
![h = 18.75 m](https://tex.z-dn.net/?f=h%20%3D%2018.75%20m)
Now when it will reach at point B then its normal force is just equal to ZERO
![N_B = 0](https://tex.z-dn.net/?f=N_B%20%3D%200)
now when it reach point C then the speed will be
![mgh - mg(2R_c) = \frac{1}{2]mv_c^2](https://tex.z-dn.net/?f=mgh%20-%20mg%282R_c%29%20%3D%20%5Cfrac%7B1%7D%7B2%5Dmv_c%5E2)
![v_c^2 = 2g(h - 2R_c)](https://tex.z-dn.net/?f=v_c%5E2%20%3D%202g%28h%20-%202R_c%29)
![v_c = 13.1 m/s](https://tex.z-dn.net/?f=v_c%20%3D%2013.1%20m%2Fs)
now normal force at point C is given as
![F_n = \frac{mv_c^2}{R_c} - mg](https://tex.z-dn.net/?f=F_n%20%3D%20%5Cfrac%7Bmv_c%5E2%7D%7BR_c%7D%20-%20mg)
![F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)](https://tex.z-dn.net/?f=F_n%20%3D%20%5Cfrac%7B700%5Ctimes%2013.1%5E2%7D%7B5%7D%20-%20%28700%20%5Ctimes%209.8%29)
![F_n = 1.72 \times 10^4](https://tex.z-dn.net/?f=F_n%20%3D%201.72%20%5Ctimes%2010%5E4)