Answer:
14778.29 N
Explanation:
Diameter, d=10.1mm= 
Since stress, 
where F is force, A is area and since specimen is cylindrical,
A= 
Therefore, 
Also strain 
where L is length
Poison’s ratio,v is the ratio of lateral strain to longitudinal strain hence
V= 
From Hooke’s law, 
Conclusively, 
F=
F= 
= 14778.29N
Therefore, required force F is 14778.29 N
Answer:
r = 71.8⁰
Explanation:
given,
refractive index of the glass 1 = 1.70
refractive index of glass 2 = 1.58
angle of incidence = 62°
angle of refraction =?
using Snell's law
1.7 ×sin 62 ^0 = 1.58× sin r
sin r = 0.95
r = sin⁻¹(0.95)
r = 71.8⁰
angle of refraction =r = 71.8⁰
Answer:
The torque on the wrench is 4.188 Nm
Explanation:
Let r = xi + yj where is the distance of the applied force to the origin.
Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,
r = 0.18i + 0.055j
The applied force f = 88i - 23j
The torque τ = r × F
So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j
= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j
= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0 since i × i = 0, j × j = 0, i × j = k and j × i = -k
= 0 - 4.14k + 0.0484(-k) + 0
= -4.14k - 0.0484k
= -4.1884k Nm
≅ -4.188k Nm
So, the torque on the wrench is 4.188 Nm
<h3><u>Answer</u>;</h3>
= 22°
<h3><u>Explanation</u>;</h3>
- According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.
- Therefore; Sin i/Sin r = η
In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33
Thus;
Sin 30 / Sin r = 1.33
Sin r = Sin 30°/1.33
= 0.3759
r = Sin^-1 0.3759
= 22.08
<u>≈ 22°</u>
Answer:
Force applied to smaller cross section is
= 82.63 N
Explanation:
As we know
where 
signifies the weight of the two chair in a hydraulic-lift system
And 
signifies the area of the two respective chairs in a hydraulic-lift system
Given -
N
Square centimeter
Square centimeter
Substituting the given values in above equation, we get -
Force applied to smaller cross section is
= 82.63 N
