Answer:
153.2 J
Explanation:
Let's first list our given parameters;
mass (m) of the block = 10 kg
which slides down ( i.e displacement) = 2 m
kinetic coefficient of friction (μk) = 0.2
In the diagram shown below; if we take an integral look at the component of force in the direction of the displacement; we have
Fcos 40°
100 (cos 40°)
76.60 N
Workdone by the friction force can now be determined as:
W =
× displacement
W = 76.60 × 2
W = 153.2 J
∴ the work done by the friction force = 153.2 J
Answer:
The correct answer is "25.03 mm".
Explanation:
Given:
Thickness of plate,
= 10 mm
On a side,
= 100 mm
Diameter hole,
= 25 mm
Coefficient of thermal expansion,
CTE = ![12\times 10^{-6} /^{\circ} C](https://tex.z-dn.net/?f=12%5Ctimes%2010%5E%7B-6%7D%20%2F%5E%7B%5Ccirc%7D%20C)
Now,
⇒ ![D_i\times (12\times 10^{-6}) \Delta \theta = \Delta D](https://tex.z-dn.net/?f=D_i%5Ctimes%20%2812%5Ctimes%2010%5E%7B-6%7D%29%20%5CDelta%20%5Ctheta%20%3D%20%5CDelta%20D)
= ![25\times 12\times 10^{-6} \Delta \theta](https://tex.z-dn.net/?f=25%5Ctimes%2012%5Ctimes%2010%5E%7B-6%7D%20%5CDelta%20%5Ctheta)
= ![3\times 10^{-4} \Delta \theta](https://tex.z-dn.net/?f=3%5Ctimes%2010%5E%7B-4%7D%20%5CDelta%20%5Ctheta)
= ![3\times 10^{-2}](https://tex.z-dn.net/?f=3%5Ctimes%2010%5E%7B-2%7D)
hence,
The final diameter of hole will be:
Answer:
I know this sounds quite deep but it is as deep as a grave
Explanation:
It's reality
Answer:
The answer is "828.75"
Explanation:
Please find the correct question:
For W21x93 BEAM,
![Z_x = 221.00 in^3 \\\\\to \frac{b_t}{2t_f} =4.53\\\\\to \frac{h}{t_w}=32.3](https://tex.z-dn.net/?f=Z_x%20%3D%20221.00%20in%5E3%20%5C%5C%5C%5C%5Cto%20%5Cfrac%7Bb_t%7D%7B2t_f%7D%20%3D4.53%5C%5C%5C%5C%5Cto%20%5Cfrac%7Bh%7D%7Bt_w%7D%3D32.3)
For A992 STREL,
![F_y= 50\ ks](https://tex.z-dn.net/?f=F_y%3D%2050%5C%20%20ks)
Check for complete section:
![\to \frac{b_t}{2t_f} =4.53 < \frac{65}{\sqrt{f_y = 9.19}}\\\\\to \frac{h}{t_w} =32.3 < \frac{640}{\sqrt{f_y = 90.5}}](https://tex.z-dn.net/?f=%5Cto%20%5Cfrac%7Bb_t%7D%7B2t_f%7D%20%3D4.53%20%3C%20%5Cfrac%7B65%7D%7B%5Csqrt%7Bf_y%20%3D%209.19%7D%7D%5C%5C%5C%5C%5Cto%20%5Cfrac%7Bh%7D%7Bt_w%7D%20%3D32.3%20%3C%20%5Cfrac%7B640%7D%7B%5Csqrt%7Bf_y%20%3D%2090.5%7D%7D)
Design the strength of beam =![\phi_b Z_x F_y\\\\](https://tex.z-dn.net/?f=%5Cphi_b%20Z_x%20F_y%5C%5C%5C%5C)
![=0.9 \times 221 \times 50\\\\=9945 \ in \ \ kips\\\\=\frac{9945}{12}\\\\= 828.75 \ft \ kips \\](https://tex.z-dn.net/?f=%3D0.9%20%5Ctimes%20221%20%5Ctimes%2050%5C%5C%5C%5C%3D9945%20%5C%20in%20%5C%20%5C%20kips%5C%5C%5C%5C%3D%5Cfrac%7B9945%7D%7B12%7D%5C%5C%5C%5C%3D%20828.75%20%5Cft%20%5C%20kips%20%5C%5C)