<span>The acrylic plastic rod is 200 mm long and 15 mm in
diameter. If an axial load of 300 N is applied to it, determine
the change in its length and the change in its diameter.
. latd = -0.0002515 (15) = -0.00377 mm
elat = -Velong = -0.4(0.0006288) = -0.0002515
d = elong L = 0.0006288 (200) = 0.126 mm
elong = s
E = 1.697(106
)
2.70(109
) = 0.0006288
s = P
A = 300 p
4 (0.015)2 = 1.697 MPa
•3–25. T</span>
Answer:The triple beam balance can measure mass to the nearest tenth of a gram (0.1 g).
Explanation:
Answer:
The speed of the other piece of the object is 13/4*v0, going to the right.
Explanation:
Assuming that no external forces acting during the explosion (as the time during which it happens is considered infinitesimal), total momentum must be conserved , same as it happens with the total mass, assuming that v₀ is much smaller than c (speed of light), so no conversion of mass into energy is done.
So, based on the information provided, we can write the following expression, assuming the positive direction to be the one to the right:
m*v₀ = -3/5*m*(v₀/2) + 2/5*m*vₓ
Solving for vₓ, we have:
vₓ = 13/4*v₀, to the right, as the value obtained is positive.
The total heat that is needed by the water to get converted to steam is calculated below through the equation,
H = (Hv)(n)
where H is the heat, Hv is the heat of vaporization (40.67 kJ/mol), and n is the number of mols. Substituting the known values,
H = (40.66 kJ/mol)(1.5 mol) = 60.99 kJ
Calculating for the time,
t = (60990 J) / (24 J/s)
t = 2541.25 s
Answer: 2541.25 s or approximately 42.35 minutes
Answer:
The question is incomplete, the complete question is "A car drives on a circular road of radius R. The distance driven by the car is given by d(t)= at^3+bt [where a and b are constants, and t in seconds will give d in meters]. In terms of a, b, and R, and when t = 3 seconds, find an expression for the magnitudes of (i) the tangential acceleration aTAN, and (ii) the radial acceleration aRAD3"
answers:
a.![18a m/s^{2}](https://tex.z-dn.net/?f=18a%20m%2Fs%5E%7B2%7D)
b. ![a_{rad}=\frac{(27a +b)^{2}}{R}](https://tex.z-dn.net/?f=a_%7Brad%7D%3D%5Cfrac%7B%2827a%20%2Bb%29%5E%7B2%7D%7D%7BR%7D)
Explanation:
First let state the mathematical expression for the tangential acceleration and the radial acceleration.
a. tangential acceleration is express as
![a_{tan}=\frac{d|v|}{dt} \\](https://tex.z-dn.net/?f=a_%7Btan%7D%3D%5Cfrac%7Bd%7Cv%7C%7D%7Bdt%7D%20%5C%5C)
since the distance is expressed as
![d=at^{3}+bt](https://tex.z-dn.net/?f=d%3Dat%5E%7B3%7D%2Bbt)
the derivative is the velocity, hence
![V(t)=\frac{dd(t)}{dt}\\V(t)=3at^{2}+b\\](https://tex.z-dn.net/?f=V%28t%29%3D%5Cfrac%7Bdd%28t%29%7D%7Bdt%7D%5C%5CV%28t%29%3D3at%5E%7B2%7D%2Bb%5C%5C)
hence when we take the drivative of the velocity we arrive at
b. the expression for the radial acceleration is expressed as
![a_{rad}=\frac{v^{2}}{r}\\a_{rad}=\frac{(3at^{2} +b)^{2}}{R}\\t=3\\a_{rad}=\frac{(27a +b)^{2}}{R}](https://tex.z-dn.net/?f=a_%7Brad%7D%3D%5Cfrac%7Bv%5E%7B2%7D%7D%7Br%7D%5C%5Ca_%7Brad%7D%3D%5Cfrac%7B%283at%5E%7B2%7D%20%2Bb%29%5E%7B2%7D%7D%7BR%7D%5C%5Ct%3D3%5C%5Ca_%7Brad%7D%3D%5Cfrac%7B%2827a%20%2Bb%29%5E%7B2%7D%7D%7BR%7D)