Answer:
Not possible
Explanation:
= longitudinal modulus of elasticity = 35 Gpa
= transverse modulus of elasticity = 5.17 Gpa
= Epoxy modulus of elasticity = 3.4 Gpa
= Volume fraction of fibre (longitudinal)
= Volume fraction of fibre (transvers)
= Modulus of elasticity of aramid fibers = 131 Gpa
Longitudinal modulus of elasticity is given by
Transverse modulus of elasticity is given by
Hence, it is not possible to produce a continuous and oriented aramid fiber.
Answer:
609547.12 Pa ≈ 6.10×10^5 Pa
Explanation:
Step 1:
Data obtained from the question. This include the following:
Force (F) = 49.8 N
Radius (r) = 0.00510 m
Pressure (P) =..?
Step 2:
Determination of the area of the head of the nail.
The head of a nail is circular in nature. Therefore, the area is given by:
Area (A) = πr²
With the above formula we can obtain the area as follow:
Radius (r) = 0.00510 m
Area (A) =?
A = πr²
A = π x (0.00510)²
A = 8.17×10^-5 m²
Therefore the area of the head of the nail is 8.17×10^-5 m²
Step 3:
Determination of the pressure exerted by the hammer.
This is illustrated below:
Force (F) = 49.8 N
Area (A) = 8.17×10^-5 m²
Pressure (P) =..?
Pressure (P) = Force (F) /Area (A)
P = F/A
P = 49.8/8.17×10^-5
P = 609547.12 N/m²
Now, we shall convert 609547.12 N/m² to Pa.
1 N/m² = 1 Pa
Therefore, 609547.12 N/m² = 609547.12 Pa.
Therefore, the pressure exerted by the hammer on the nail is 609547.12 Pa or 6.10×10^5 Pa
Answer:
not work
Explanation:
in a series circuit, everything meaning the electrons are flowing on one path, therefore, it wouldn continue to work.
D=s(t) so it would be d=10(.19) d=.19 FOR BITH SNDWERS
Answer:
d) 1/32 microgram
Explanation:
First half life is the time at which the concentration of the reactant reduced to half.
Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.
Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.
Forth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.
Fifth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/32.
The initial mass of the sample = 1 microgram
After 5 half-lives, the mass should reduce to 1/32 of the original.
So the concentration left = 1/32 of 1 microgram = 1/32 microgram
