Answer:
F = 19.375 x 10^-6 N
Explanation:
This problem can be solved by applying Coulomb's Law, which lets us determine the force between two electrically charged particles.
It is defined as
F = (ke * q1 * q2)/ r^2
Where,
ke = is Coulomb's constant ≈ 9×10^9 N⋅m^2⋅C^−2
q1 = 5.0 x 10^-8 C
q2 = 1.0 x 10^-7 C
r = 5 ft = 1,524 m
F = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)
F = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)
F = 19.375 x 10^-6 N
<span>With a half-life of 700 million years, U-235 would have had twice as much mass at a time 700 MYA. This would have put the mass at 100kg at that time. Going back another 50 million years would be (50/700) or 1/14 of the half-life, or (1/2 * 1/14), or 1/28 of the total mass. 1/28 of 100kg is 3.57kg, so the amount present at the 750MYA mark would be approximately 103.57kg of U-235.</span>
Answer:
1.) 6 hectokisses 50 times
2.) Millionaire
Explanation:
Answer:
Fp = 26.59[N]
Explanation:
This problem can be solved using the principle of work and energy conservation, i.e. the final kinetic energy of a body will be equal to the sum of the forces that do work on the body plus the initial kinetic energy.
We need to identify the initial data:
d = distance = 41.9[m]
Ff = friction force = 44.5 [N]
m = mass = 16.3 [kg]
v1 = 1.9 [m/s]
v2 = 12.6 [m/s]
The kinetic energy at the beginning can be calculated as follows:
![E_{k1}= \frac{1}{2}*m*v_{1}^2 \\E_{k1}= \frac{1}{2}*16.3*(1.9)_{1}^2\\E_{k1}= 29.42[J]](https://tex.z-dn.net/?f=E_%7Bk1%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2Am%2Av_%7B1%7D%5E2%20%5C%5CE_%7Bk1%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2A16.3%2A%281.9%29_%7B1%7D%5E2%5C%5CE_%7Bk1%7D%3D%2029.42%5BJ%5D)
And the final kinetic energy.
![E_{k2}= \frac{1}{2}*m*v_{2}^2 \\E_{k2}= \frac{1}{2}*16.3*(12.6)^2\\E_{k2}= 1294[J]](https://tex.z-dn.net/?f=E_%7Bk2%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2Am%2Av_%7B2%7D%5E2%20%5C%5CE_%7Bk2%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2A16.3%2A%2812.6%29%5E2%5C%5CE_%7Bk2%7D%3D%201294%5BJ%5D)
The work is performed by two forces, the friction force and the pushing force, it is important to clarify that these forces are opposite in direction.
The weight of the cart also performs a work in the direction of movement since the plane is tilted down, this component of the weight of the cart must be parallel to the surface of the inclined plane.
![W_{1-2}=-(44.5*41.9)+(16.3*9.81*sin(17.5)*41.9)+(F_{p}*41.9) \\therefore:\\E_{k1}+W_{1-2}=E_{k2}\\29.42+150.16+(F_{p}*41.9)=1294\\F_{p}=1114.42/41.9\\F_{p}=26.59[N]](https://tex.z-dn.net/?f=W_%7B1-2%7D%3D-%2844.5%2A41.9%29%2B%2816.3%2A9.81%2Asin%2817.5%29%2A41.9%29%2B%28F_%7Bp%7D%2A41.9%29%20%5C%5Ctherefore%3A%5C%5CE_%7Bk1%7D%2BW_%7B1-2%7D%3DE_%7Bk2%7D%5C%5C29.42%2B150.16%2B%28F_%7Bp%7D%2A41.9%29%3D1294%5C%5CF_%7Bp%7D%3D1114.42%2F41.9%5C%5CF_%7Bp%7D%3D26.59%5BN%5D)
The index of refraction of a material is the ratio between the speed of light in vacuum, c, and the speed of light in that material, v:

where the speed of light in vacuum is

. The speed of light in benzene is

, so we can use the previous relationship to find the refractive index of benzene: