P1 P3 P₁ -5m B B B 3 m 3 m 3 m I=200000 cm³ Ib-160000 cm²* E=30000 MPa m=25 tons

Q3/ If the section of the Dam is as shown below. What is the area of the section in (m²), if the scale was (1:250). 50 cm 40 cm 150 cm

Q3/ If the section of the Dam is as shown below. What is the area of the section in (m²), if the scale was (1:250). 20 cm 15 cm 12 cm 10 cm

Refer to Figure 03.24. Given that H, = 300 mm, H2 = 500 mm, and ly - 600 mm and that at z = 200 mm, l1 - 500 mm. It is required to determine li at z = 600 mm. The following general equations may be useful: kre=H/[(H/k1) + (Hz/kız) + ... + (H./k..)] - (1/H) kw H, + ki2 H2 + ... + k H.) a) What is the ratio ki/ky? b) What is the value of h at z = 600 mm? c) If k, = 5 x 106 cm/s, what is the equivalent k of soils 1 and 2? Water supply B Figure 03.24

The object W1 in the figure is at an angle of 45 to the horizontalIt is based on its frictionless surface at point B.Point A is articulated (fixed support) and ground frictionSince (μ = 0.3) body W2 is a surfaceWhat should be the weight of W1 that will cause it to start? W2= 38 [kN]sin45=cos45=0.7

1. Given the velocity field: V = (6 + 2xy + *)i – (xy² + 101) j + 25k. What is the acceleration of a particle at (3, 0, 2) at time i = 1? [IIT Bombay] [Ans. + 58.35 units]

Dimension the cross section of the rectangular cross section cantilever bar given loading status and dimensions. (Ostetx = 1200 N/cm²; Listetx = 180 N/cm) 4000N b . 4 N/cm h = 2b 400 cm

PROBLEM 2 The two forces are equivalent to a force R that has a line of action passing through point A. Determine R and the distance X. ķ 200mm-f 100 N N X 90mm

As a train accelerates uniformly it passes successive 800 meter marks while traveling at velocities of 3 m/s and then 12 m/s. [Select] ✓(a) For the first mark of the journey, what is the acceleration of the train in m/s². [Select] [Select] (b) Determine the train's velocity when it passes the next mark in m/s. (c) Determine the time it takes to travel the 1.6-km distance in seconds.

4. Solve the following system of equations. 8 ft 2 ft 64 BD- 12 12 0 = ЕС 4 ft 7 32 BD+° EC– 1080 12 18 0 = A 3 ft 7 E В 6 ft 2 ft 5 ft