Answer:
Explanation:
Let the critical angle be C .
sinC = 1 / μ where μ is index of refraction .
sinC = 1 /1.2
= .833
C = 56°
Then angle of refraction r = 90 - 56 = 34 ( see the image in attached file )
sin i / sinr = 1.2 , i is angle of incidence
sini = 1.2 x sinr = 1.2 x sin 34 = .67
i = 42°.
Answer:
(a) T= 38.4 N
(b) m= 26.67 kg
Explanation:
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Kinematics
d= v₀t+ (1/2)*a*t² (Formula 2)
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
v₀=0, d=18 m , t=5 s
We apply the formula 2 to calculate the accelerations of the blocks:
d= v₀t+ (1/2)*a*t²
18= 0+ (1/2)*a*(5)²
a= (2*18) / ( 25) = 1.44 m/s²
to the right
We apply Newton's second law to the block A
∑Fx = m*ax
60-T = 15*1.44
60 - 15*1.44 = T
T = 38.4 N
We apply Newton's second law to the block B
∑Fx = m*ax
T = m*ax
38.4 = m*1.44
m= (38.4) / (1.44)
m = 26.67 kg
Answer:
W = 55.12 J
Explanation:
Given,
Natural length = 6 in
Force = 4 lb, stretched length = 8.4 in
We know,
F = k x
k is spring constant
4 = k (8.4-6)
k = 1.67 lb/in
Work done to stretch the spring to 10.1 in.

![W = \dfrac{k}{2}[x^2]_6^{10.1}](https://tex.z-dn.net/?f=W%20%3D%20%5Cdfrac%7Bk%7D%7B2%7D%5Bx%5E2%5D_6%5E%7B10.1%7D)

W = 55.12 J
Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.
Answer:
the solid particles that settle down at the bottom of the container are known as sediment.
Explanation: