Answer:
![d=0.165m](https://tex.z-dn.net/?f=d%3D0.165m)
Explanation:
Given
,
,
,![v=2\frac{rev}{s}](https://tex.z-dn.net/?f=v%3D2%5Cfrac%7Brev%7D%7Bs%7D)
The tension of the spring is
![F_{k}=K*x_{1}=m*g](https://tex.z-dn.net/?f=F_%7Bk%7D%3DK%2Ax_%7B1%7D%3Dm%2Ag)
![K=\frac{m*g}{x_{1}}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7Bm%2Ag%7D%7Bx_%7B1%7D%7D)
![K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B0.25kg%2A9.8m%2Fs%5E2%7D%7B0.05m%7D%3D49N%2Fm)
The force in the spring is equal to centripetal force so
![F_{c}=\frac{m*v^2}{r}](https://tex.z-dn.net/?f=F_%7Bc%7D%3D%5Cfrac%7Bm%2Av%5E2%7D%7Br%7D)
![v=w*r=2\pi*r](https://tex.z-dn.net/?f=v%3Dw%2Ar%3D2%5Cpi%2Ar)
But Fc is also
Fc=KxΔr
![F_{c}=K*(r-x_{2})](https://tex.z-dn.net/?f=F_%7Bc%7D%3DK%2A%28r-x_%7B2%7D%29)
Replacing
![m*4\pi^2*r=K*(r-x_{2})](https://tex.z-dn.net/?f=m%2A4%5Cpi%5E2%2Ar%3DK%2A%28r-x_%7B2%7D%29)
![0.25kg*4\pi^2*r=49*(r-0.04m)](https://tex.z-dn.net/?f=0.25kg%2A4%5Cpi%5E2%2Ar%3D49%2A%28r-0.04m%29)
![r=0.205m](https://tex.z-dn.net/?f=r%3D0.205m)
total distance is
![d=0.205-0.04=0.165m](https://tex.z-dn.net/?f=d%3D0.205-0.04%3D0.165m)
Detailed Explanation:
1) Rusting of Iron
4Fe + 3O2 + 2H2O -> 2Fe2O32H2O
Reactants :-
Fe = 4
O = 3 * 2 + 2 = 8
H = 2 * 2 = 4
Products :-
Fe = 2 * 2 = 4
O = 2 * 3 + 2 = 8
H = 2 * 2 = 4
2) Fermentation of sucrose…
C12H22O11 + H2O -> 4C2H5OH + 4CO2
Reactants :-
C = 12
H = 22 + 2 = 24
O = 11 + 1 = 12
Products :-
C = 4 * 2 + 4 = 12
H = 4 * 5 + 4 = 24
O = 4 * 2 + 4 = 12
Looking closely at the way I have taken the total number of elements on the reactants and products side, you can solve the rest.
All the Best!
Answer:
0.29 m
Explanation:
9 mm = 0.009 m in diameter
Cross-sectional area ![A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20d%5E2%2F4%20%3D%20%5Cpi%20%2A%200.009%5E2%2F4%20%3D%206.36%5Ctimes%2010%5E%7B-5%7D%20m%5E2)
Let the tensile modulus of Nickel
.
The elongation of the rod can be calculated using the following formula:
![\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m](https://tex.z-dn.net/?f=%5CDelta%20L%20%3D%20%5Cfrac%7BF%20L%7D%7BA%20E%7D%20%3D%20%5Cfrac%7B6283%2A50%7D%7B6.36%5Ctimes%2010%5E%7B-5%7D%20%2A%20170%20%5Ctimes%2010%5E9%7D%20%3D%20%5Cfrac%7B314150%7D%7B1081200%7D%20%3D%200.29%20m)
Answer: If all forces acting on a car are zero, than the cars speed is zero since there are no forces to push or pull the car :)
Explanation: