Answer:
yes
Explanation:
He used the formula R = VI and obtained an
answer of 2
<span>The speed of longitudinal waves, S, in a thin rod = âšYoung modulus / density , where Y is in N/m^2.
So, S = âšYoung modulus/ density. Squaring both sides, we have, S^2 = Young Modulus/ density.
So, Young Modulus = S^2 * density; where S is the speed of the longitudinal wave.
Then Substiting into the eqn we have (5.1 *10^3)^2 * 2.7 * 10^3 = 26.01 * 10^6 * 2.7 *10^6 = 26.01 * 2.7 * 10^ (6+3) = 70.227 * 10 ^9</span>
Answer:
The angle between the blue beam and the red beam in the acrylic block is
Explanation:
From the question we are told that
The refractive index of the transparent acrylic plastic for blue light is 
The wavelength of the blue light is 
The refractive index of the transparent acrylic plastic for red light is 
The wavelength of the red light is 
The incidence angle is 
Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as
![r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ]](https://tex.z-dn.net/?f=r_F%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%28i%29%20%2A%20%20n_a%20%7D%7Bn_F%7D%20%5D)
Where 
is the refractive index of air which have a value of
So
![r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ]](https://tex.z-dn.net/?f=r_F%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%2845%29%20%2A%20%201%20%7D%7B%201.497%7D%20%5D)
Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as
![r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ]](https://tex.z-dn.net/?f=r_C%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%28i%29%20%2A%20%20n_a%20%7D%7Bn_C%7D%20%5D)
Where is the refractive index of air which have a value of
So
![r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ]](https://tex.z-dn.net/?f=r_C%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%2845%29%20%2A%20%201%20%7D%7B%201.488%7D%20%5D)
The angle between the blue beam and the red beam in the acrylic block
substituting values
1) 15 / 12 = 1.25 ratio
2) to increase acceleration 1.25 times (with same F, or same engine) you have to lower mass 1.25 times
3) 1515/1.25 = 1212 kg
choose A
Answer: 35*10^3 N/m
Explanation: In order to explain this problem we know that the potential energy for spring is given by:
Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.
We also know that with additional streching of 2 cm of teh spring, the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.
Then the difference of energy for both cases is 7 J so:
ΔUp= 1/2*k* (0.02)^2 then
k=2*7/(0.02)^2=35000 N/m
