Answer:
![15^{\circ}](https://tex.z-dn.net/?f=15%5E%7B%5Ccirc%7D)
Explanation:
Using Snell's law which is represented by
![n_1sin\theta_1 = n_2sin\theta_2](https://tex.z-dn.net/?f=n_1sin%5Ctheta_1%20%3D%20n_2sin%5Ctheta_2)
Making
the subject of the formula then
![\theta_2=sin^{-1}(\frac {n_1sin\theta_1}{n_2})](https://tex.z-dn.net/?f=%5Ctheta_2%3Dsin%5E%7B-1%7D%28%5Cfrac%20%7Bn_1sin%5Ctheta_1%7D%7Bn_2%7D%29)
Here
and
are the angles of incidence and refraction in water and air respectively
and
are refraction index
Substituting 1.0003 for
and 1.33 for
then
for
we obtain
![\theta_2=sin^{-1}(\frac {1.0003\times sin 20^{\circ}}{1.33})=14.90606875^{\circ}\approx 15^{\circ}](https://tex.z-dn.net/?f=%5Ctheta_2%3Dsin%5E%7B-1%7D%28%5Cfrac%20%7B1.0003%5Ctimes%20sin%2020%5E%7B%5Ccirc%7D%7D%7B1.33%7D%29%3D14.90606875%5E%7B%5Ccirc%7D%5Capprox%2015%5E%7B%5Ccirc%7D)
Answer:
v = 3.08 m/s.
Explanation:
Given,
- Initial velocity of the block = u = 43 cm/s = 0.43 m/s
- mass of the block = m = 1.40 kg
- spring constant = k = 15.5 N/m.
- x = 0.650 m
At x = 0.650 m
Let 'a' be the acceleration of the block at x.
Total force due to the spring force on the blcok at x= kx
![\therefore F_s\ =\ kx\\\Rightarrow ma\ =\ kx\\\Rightarrow a\ =\ \dfrac{kx}{m}\\\Rightarrow a\ =\ \dfrac{15.5\times 0.650}{1.40}\\\Rightarrow a\ =\ 7.19\ m/s^2](https://tex.z-dn.net/?f=%5Ctherefore%20F_s%5C%20%3D%5C%20kx%5C%5C%5CRightarrow%20ma%5C%20%3D%5C%20kx%5C%5C%5CRightarrow%20a%5C%20%3D%5C%20%5Cdfrac%7Bkx%7D%7Bm%7D%5C%5C%5CRightarrow%20a%5C%20%3D%5C%20%5Cdfrac%7B15.5%5Ctimes%200.650%7D%7B1.40%7D%5C%5C%5CRightarrow%20a%5C%20%3D%5C%207.19%5C%20m%2Fs%5E2)
Hence the acceleration of the block at x is 7.19\ m/s^2.
Let 'v' be the velocity of the block at x.
From the kinematics,
![v^2\ =\ u^2\ +\ 2ax\\\Rightarrow v\ =\ \sqrt{u^2\ +\ 2ax}\\\Rightarrow v\ =\ \sqrt{0.43^2\ +\ 2\times 7.19\times 0.650}\\\Rightarrow v\ =\ 3.08\ m/s.](https://tex.z-dn.net/?f=v%5E2%5C%20%3D%5C%20u%5E2%5C%20%2B%5C%202ax%5C%5C%5CRightarrow%20v%5C%20%3D%5C%20%5Csqrt%7Bu%5E2%5C%20%2B%5C%202ax%7D%5C%5C%5CRightarrow%20v%5C%20%3D%5C%20%5Csqrt%7B0.43%5E2%5C%20%2B%5C%202%5Ctimes%207.19%5Ctimes%200.650%7D%5C%5C%5CRightarrow%20v%5C%20%3D%5C%203.08%5C%20m%2Fs.)
Hence at the compression x = 0.650 m the velocity of the block is 3.08 m/s.
A: Human Body
C is wrong because they don’t have the tools to test it on another planet
Answer:
a) r eq = -a/(2b)
b) k = a/r eq = -2b
Explanation:
since
U(r) = ar + br²
a) the equilibrium position dU/dr = 0
U(r) = a + 2br = 0 → r eq= -a/2b
b) the Taylor expansion around the equilibrium position is
U(r) = U(r eq) + ∑ Un(r eq) (r- r eq)^n / n!
,where Un(a) is the nth derivative of U respect with r , evaluated in a
Since the 3rd and higher order derivatives are =0 , we can expand until the second derivative
U(r) = U(r eq) + dU/dr(r eq) (r- r eq) + d²U/dr²(r eq) (r- r eq)² /2
since dU/dr(r eq)=0
U(r) = U(r eq) + d²U/dr²(r eq) (r- r eq)² /2
comparing with an energy balance of a spring around its equilibrium position
U(r) - U(r eq) = 1/2 k (r-r eq)² → U(r) = U(r eq) + 1/2 k (r-r eq)²
therefore we can conclude
k = d²U/dr²(r eq) = -2b , and since r eq = -a/2b → -2b=a/r eq
thus
k= a/r eq
Answer:
Mass = Size, Weight = Heaviness
Explanation:
Mass is how much matter that is in an object
Weight is a measurement that indicates how heavy something