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Let u, v, and w be distinct vectors of a real vector space V . [6 marks] (a). Show that span{u, v, w}=span{u + v − w, u + w − v, v + w −

by Admin | Oct 19, 2024 | Questions

Question: Let u, v, and w be distinct vectors of a real vector space V . [6 marks] (a). Show that span{u, v, w}=span{u + v − w, u + w − v, v + w − Answer: To show that ( \text{span}{u, v, w} = \text{span}{u + v – w, u + w – v, v + w – u} ), we need...

The curvilinear motion of a particle is defined by v, = 50 – 16t and y =100 – 4t’, where v, is in meters per second, y is in meters, and t is in seconds.It is also known that x = 0 when t – 0. Plot the path of the particle and determine its velocity and acceleration when the position y = 0 is reached.

by Admin | Jul 30, 2024 | Questions

To plot the path of the particle, we need to find the expression for x in terms of t. We know that the velocity v_x is given by: vx=dxdt=50−16t𝑣𝑥=𝑑𝑥𝑑𝑡=50−16𝑡 Integrating with respect to t, we get: ∫dx=∫(50−16t)dt∫𝑑𝑥=∫(50−16𝑡)𝑑𝑡 x=50t−8t2+C𝑥=50𝑡−8𝑡2+𝐶 Given that x = 0...

A motor drives two loads. One has rotational motion. It is coupled to the motor through a reduction gear with a = 0.1 and efficiency of 90%. The load has a moment of inertia of 10.kg- m ^ 2 and a torque of 10 N-m. Other load has translational motion and consists of 1000 kg weight to be lifted up at an uniform speed of 1.5 m/s. Coupling between this load and the motor has an efficiency of 85%. Motor has an inertia of 0.2kg – m ^ 2 and runs at a constant speed of 1420 rpm. Determine equivalent inertia referred to the shaft and power developed by the motor.

by Admin | Jul 29, 2024 | Questions

• If the motor is rated at 19kW, is the motor sufficient to drive the two loads? • The translational motion load now has to lift a weight of 1200 kg at the same speed of 1.5m/s. Is the motor still capable to drive both loads at the same motor speed of 1420 rpm? Answer...

For the beams of problems 6.2-6.16, draw the shear force and bending moment diagrams and find the maximum shear force, maximum bending moment and point(s) of contraflexure (PCF).ranscribed Image Text:6.9 5 kN 2 kN 3 kN 4 kN 25 kN/m 25 kN/m 10 kN/m 2 m 2 m 2m 1 m 1 m Figure 6.38 (R, = 69,6 kN; R = 60,4 kN; Vmax = 44,6 kN, B; Mmax = -30 kN.m, B; PCF 1,75 m from F and 2,90 m from AJ

by Admin | Jul 29, 2024 | Questions

Answer: To analyze the beam described in the problem and to draw the shear force and bending moment diagrams, follow these steps: Given Data Loads: 5 kN 2 kN 3 kN 4 kN 25 kN/m (Uniformly Distributed Load, UDL) for 2 m 25 kN/m (UDL) for 2 m 10 kN/m (UDL) for 1 m Beam...