Answer:
A = 5h (B + b); solve for B.
2
Answer:
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Explanation:
Answer:
The strength coefficient is
and the strain-hardening exponent is 
Explanation:
Given the true strain is 0.12 at 250 MPa stress.
Also, at 350 MPa the strain is 0.26.
We need to find
and the
.

We will plug the values in the formula.

We will solve these equation.
plug this value in 

Taking a natural log both sides we get.

Now, we will find value of 

So, the strength coefficient is
and the strain-hardening exponent is
.
Given that,
Mass of the object 1, m = 107.01 grams
To find,
Force on the object.
Solution,
The force acting on the object is gravitational force. The force is given by the formula as follow :
F = mg
g is acceleration due to gravity
F = 0.10701 kg × 9.8 m/s²
F = 1.048 N
So, the force acting on object 1 is 1.048 N.
Answer:

Solution:
Note: Refer the diagram


Drag coefficient data for selected objects table at
Hemisphere (open end facing flow), 
Substituting all parameters,

Then,
![\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26V_%7Bb%7D%3DV_%7Bw%7D-%5Cleft%5B%5Cfrac%7B2%20F_%7BR%7D%7D%7B%5Crho%5Cleft%28C_%7BD%2C%20w%7D%20A_%7Bw%7D%2BC_%7BD%2C%20B%7D%20A_%7Bb%7D%5Cright%29%7D%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cdots%5C%5C%26V_%7Bw%7D%3D24%20%5Ctimes%201000%20%5Ctimes%20%5Cfrac%7B1%7D%7B3600%7D%5C%5C%26V_%7Bw%7D%3D6.67%20%5Cfrac%7B%20m%20%7D%7B%20s%20%7D%5Cend%7Baligned%7D)
And the equation becomes,
![\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26V_%7Bb%7D%3D6.67-%5Cleft%5B%5Cfrac%7B2%20%5Ctimes%205.52%7D%7B1.23%281.42%20%5Ctimes%201.17%2B1.2%20%5Ctimes%200.3%29%7D%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%26V_%7Bb%7D%3D6.67-2.11%5C%5C%26V_%7Bb%7D%3D4.56%20%5Cfrac%7B%20m%20%7D%7B%20s%20%7D%5Cend%7Baligned%7D)
Thus the floyds travels at
wind speed.