Answer:
3. 9600m/s
4. 2.5m/s
Explanation:
To find the final velocity of the object which accelerates at 40m/s/s for four minutes, we first need to convert the 4 minutes to seconds. There are 60 seconds in a minute so 4 minutes is equal to 4x60 = 240 seconds.
To calculate the final velocity we just multiply the acceleration rate by the time, so 40x240=9,600 m/s
The final velocity of the object would be 9600m/s.
If an object's final velocity was 115m/s after 45 seconds then we need to find out how much it accelerated in that 45 seconds to find the starting velocity. 45x2.5=112.5 It increased in speed by 112.5m/s over 45 seconds, so its initial speed was 112.5m/s slower than its final speed. 115-112.5=2.5, the object's starting velocity was 2.5m/s.
Hope this helped!
Compute the work done on the table:
<em>W</em> = <em>Fd</em> = (320 N) (32 m) = 10,240 J
Divide this by the given time duration to get the power output:
<em>P</em> = <em>W</em>/∆<em>t</em> = (10,240 J) / (150 s) ≈ 63.3 W
Well, the 20 seconds gives us enough information to actually calculate the power she has on her feet :)
Well, the work is 500*15=7500J, that is 7.5 kJ or about 18 calories (1 cal is 418J?) The power is that 7.5 kJ divided by 20 seconds: 375W. That's... OK, I suppose...
Answer:
Explanation:
Given that:
width b=100mm
depth h=150 mm
length L=2 m =200mm
point load P =500 N
Calculate moment of inertia
Point C is subjected to bending moment
Calculate the bending moment of point C
M = P x 1.5
= 500 x 1.5
= 750 N.m
M = 750 × 10³ N.mm
Calculate bending stress at point C
Calculate the first moment of area below point C
Now calculate shear stress at point C
Calculate the principal stress at point C
![\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa](https://tex.z-dn.net/?f=%5Csigma_%7B1%2C2%7D%3D%5Cfrac%7B%5Csigma_x%2B%5Csigma_y%7D%7B2%7D%20%5Cpm%5Csqrt%7B%28%5Cfrac%7B%5Csigma_x-%5Csigma_y%7D%7B2%7D%20%29%20%2B%20%28%5Ctau%29%5E2%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B666.67%2B0%7D%7B2%7D%20%5Cpm%5Csqrt%7B%28%5Cfrac%7B666.67-0%7D%7B2%7D%20%29%5E2%20%5Cpm%2844.44%29%5E2%7D%20%5C%20%5B%20%5Csigma_y%3D0%5D%5C%5C%5C%5C%3D333.33%5Cpm336.28%5C%5C%5C%5C%20%5Csigma_1%3D333.33%2B336.28%5C%5C%3D669.61KPa%5C%5C%5C%5C%5Csigma_2%3D333.33-336.28%5C%5C%3D-2.95KPa)
Calculate the maximum shear stress at piont C
