Answer:
1e , 2j , 3c , 4d , 5k , 6a , 7i , 8b , 9m , 10h ,11g , 12f , 13l .
A) -3.75 meters/second
A=(20^2-80^2)/(2x800)
=(400-6400)/1600
=-6000/1600
=-3.75
B) 16 seconds
t=(20-80)/-3.75
=-60/-3.75
=16
Answer: thickness h = 0.014cm
Question: In the manufacturing of computer chips, cylinders of silicon are cut into thin wafers that are 3.30 inches in diameter and have a mass of 1.50 g of silicon. How thick (mm) is each wafer if silicon has a density of 2.33 g/cm 3 ? (The volume of a cylinder is V=πr 2 h )
Explanation:
The volume of a cylinder is
Volume V = πr^2h ....1
The density of a material is
Density D = mass m / volume V
D = m/V ....2
Since m and D are given, we can make V the subject of formula.
V = m/D ....3
From equation 1, we need to derive the thickness h of the cylindrical silicon.
h = V/πr^2 .....4
Substituting equation 3 into 4
h = (m/D)/πr^2 .....5
Given.
mass m = 1.50g
density D = 2.33g/cm^3
radius r = diameter/2 = 3.00in/2 = 7.62/2 cm = 3.81cm
Substituting the given values into the equation
h = (1.5/2.33)/(π ×3.81^2)
thickness h = 0.014cm
The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
<h3>Force exerted outside the wheel</h3>
The force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.
∑τ = 0
- Let the distance traveled by the load = 1.5 m
- Let the radius of the wheel or position of the force = 0.45 m
∑τ = R(mg) - r(F)
rF = R(mg)
0.45F = 1.5(21,200 x 9.8)
F = 6.925 x 10⁵ N.
Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
Learn more about angular momentum here: brainly.com/question/7538238
