Answer:
Expression: 66(19+14)
Answer: 2178
Step-by-step explanation:
<u>66(19+14)</u>
66(33)
2178
Answer:
Step-by-step explanation:
The model fo the shell is given by the following equation of equilibrium:
This first-order differential equation has separable variables, which are cleared herein:
The solution of this integral is:
![t = -\frac{m}{2b}\cdot \left[\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } }\right) - \tan^{-1} \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }\right)\right]](https://tex.z-dn.net/?f=t%20%3D%20-%5Cfrac%7Bm%7D%7B2b%7D%5Ccdot%20%5Cleft%5B%5Ctan%5E%7B-1%7D%20%5Cleft%28%5Cfrac%7Bv%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%5Cright%29%20-%20%5Ctan%5E%7B-1%7D%20%5Cleft%28%5Cfrac%7B600%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%5Cright%29%5Cright%5D)
![\frac{v}{\sqrt{\frac{m\cdot g}{b} } }=\tan \left[-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\right]](https://tex.z-dn.net/?f=%5Cfrac%7Bv%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%3D%5Ctan%20%5Cleft%5B-%5Cfrac%7B2%5Ccdot%20b%5Ccdot%20t%7D%7Bm%7D%20%2B%20%5Ctan%5E%7B-1%7D%5Cleft%28%5Cfrac%7B600%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%20%20%5Cright%29%5Cright%5D)
![v = \sqrt{\frac{m\cdot g}{b} } \left [\frac{\tan \left(-\frac{2\cdot b \cdot t}{m} \right)+ \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)}{1 - \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\cdot \tan \left(-\frac{2\cdot b \cdot t}{m} \right) }\right]](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%5Cleft%20%5B%5Cfrac%7B%5Ctan%20%5Cleft%28-%5Cfrac%7B2%5Ccdot%20b%20%5Ccdot%20t%7D%7Bm%7D%20%20%5Cright%29%2B%20%5Cleft%28%5Cfrac%7B600%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%20%20%5Cright%29%7D%7B1%20-%20%5Cleft%28%5Cfrac%7B600%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%20%20%5Cright%29%5Ccdot%20%5Ctan%20%5Cleft%28-%5Cfrac%7B2%5Ccdot%20b%20%5Ccdot%20t%7D%7Bm%7D%20%20%5Cright%29%20%7D%5Cright%5D)
Answer:
1.544*10⁹ Linebackers would be required in order to obtain the same density as an alpha particle
Step-by-step explanation:
Assuming that the pea is spherical ( with radius R= 0.5 cm= 0.005 m), then its volume is
V= 4/3π*R³ = 4/3π*R³ = 4/3*π*(0.005 m)³ = 5.236*10⁻⁷ m³
the mass in that volume would be m= N*L (L= mass of linebackers=250Lbs= 113.398 Kg)
The density of an alpha particle is ρa= 3.345*10¹⁷ kg/m³ and the density in the pea ρ will be
ρ= m/V
since both should be equal ρ=ρa , then
ρa= m/V =N*L/V → N =ρa*V/L
replacing values
N =ρa*V/L = 3.345*10¹⁷ kg/m³ * 5.236*10⁻⁷ m³ /113.398 Kg = 1.544*10⁹ Linebackers
N=1.544*10⁹ Linebackers
Answer:
3.25
Step-by-step explanation:
I plugged in 5 for x and got 3.25 when I solved
