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h = 0 when the ball hit the ground, so
<span>0 = rt - 16t^2 </span>
<span>Since r = 32, </span>
<span>0= 32t - 16t^2 </span>
<span>16t(2 - t) = 0 </span>
<span>t = 0 or 2 </span>
<span>At t=0, h=0 but the ball hasn't left the ground yet. </span>
<span>Thus, the answer is 2.</span>
Answer:
E = 1580594.95 N/C
Explanation:
To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:
(1)
dS: differential of the Gaussian surface
Qin: charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2
The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:
(2)
Qin is calculate by using the charge density:
(3)
Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.
The charge density is given by:
![\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%7BQ%7D%7BV%7D%3D%5Cfrac%7B13%2A10%5E%7B-6%7DC%7D%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%28%280.15m%29%5E3-%280.10m%29%5E3%29%7D%5C%5C%5C%5C%5Crho%3D1.30%2A10%5E%7B-3%7D%5Cfrac%7BC%7D%7Bm%5E3%7D)
Next, you use the results of (3), (2) and (1):
![E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})](https://tex.z-dn.net/?f=E%284%5Cpi%20r%5E2%29%3D%5Cfrac%7B4%7D%7B3%5Cepsilon_o%7D%28r%5E3-a%5E3%29%5Crho%5C%5C%5C%5CE%3D%5Cfrac%7B%5Crho%7D%7B3%5Cepsilo_o%7D%28r-%5Cfrac%7Ba%5E3%7D%7Br%5E2%7D%29)
Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:
![E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1.30%2A10%5E%7B-3%7DC%2Fm%5E3%7D%7B3%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%28%280.112m%29-%5Cfrac%7B%280.10%29%5E3%7D%7B%280.112m%29%5E2%7D%29%5C%5C%5C%5CE%3D1%2C580%2C594.95%5Cfrac%7BN%7D%7BC%7D)
hence, the electric field is 1580594.95 N/C
Answer:
The total momentum is ![p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s](https://tex.z-dn.net/?f=p__%7BT%20%7D%7D%20%3D%282400%20%20-4%20v_2%29%20%5C%20Dalton%20%5Ccdot%20m%2Fs)
Explanation:
The diagram illustration this system is shown on the first uploaded image (From physics animation)
From the question we are told that
The mass of the first object is ![M_1 = 12 \ Dalton](https://tex.z-dn.net/?f=M_1%20%3D%20%2012%20%5C%20Dalton)
The speed of the first mass is ![v_1 = 200 \ m/s](https://tex.z-dn.net/?f=v_1%20%3D%20%20200%20%5C%20m%2Fs)
The mass of the second object is ![M_2 = 4 \ Dalton](https://tex.z-dn.net/?f=M_2%20%3D%20%204%20%5C%20Dalton)
The speed of the second object is assumed to be ![- v_2](https://tex.z-dn.net/?f=-%20v_2)
The total momentum of the system is the combined momentum of both object which is mathematically represented as
![p__{T }} = M_1 v_1 + M_2 v_2](https://tex.z-dn.net/?f=p__%7BT%20%7D%7D%20%3D%20M_1%20v_1%20%2B%20M_2%20v_2)
substituting values
![p__{T }} = 12 * 200 + 4 * (-v_2)](https://tex.z-dn.net/?f=p__%7BT%20%7D%7D%20%3D%2012%20%2A%20200%20%2B%204%20%2A%20%28-v_2%29)
![p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s](https://tex.z-dn.net/?f=p__%7BT%20%7D%7D%20%3D%282400%20%20-4%20v_2%29%20%5C%20Dalton%20%5Ccdot%20m%2Fs)
I'm pretty sure you can find it out by using a speed monitor and compass or just observing it
if there any answer choices tell me.
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HOPE I HELPED!!!</h2>
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