“This campus is a hidden gem and I encourage anyone who has the chance to make a visit to do so,” Ms Cleeland said. Ms Cleeland encouraged everyone in the region to understand the benefits of having this university in the Goulburn Valley. “Additionally, the NorVicFoods Agri-food Innovation Cluster is undertaking a broad range of research to increase efficiency in the food and agriculture sectors.” “The Victoria Drought Resilience Adoption and Innovation Hub located here is supporting drought preparedness and providing high-level advice to regional communities. “The team showed us the laboratories, teaching spaces and accommodation that support students from Melbourne, overseas and regional Victoria. “Having this wonderful campus in our own backyard really is a great asset for the region,” Ms Cleeland said. (266.State Member for Euroa Annabelle Cleeland said the university was engaged in several important projects for the future of the region. How much further will the train move before coming to rest? (Assuming the retardation to be constant). The maximum initial velocity of the ball = V i =?Ģ.9 When brakes are applied, the speed of a train decreases from 96 kmh-1 to 48 kmh-1 in 800 m.Maximum height reached by the ball S = h =?.Velocity at maximum height = V f = 0 ms -1 Time to reach maximum height (one sided time) = t = 6/2 = 3 s Solution: Acceleration due to gravity = g = -10 ms -1 (for upward motion) Calculate (i) Maximum height reached by the ball (ii) initial velocity of the ball (45m, 30 ms-1) Find the total distance travelled by train.įinal velocity = V f = 48 kmh -1 = 48 x 1000/3600 = 13.333 ms -1Ģ.8 A cricket ball is hit vertically upwards and returns to ground 6 s later. Finally, it moves with uniform retardation and is stopped after 3 minutes. (36 kmh -1)Ģ.7 A train starting from rest, accelerates uniformly and attains a velocity of 48 kmh -1 in 2 minutes. Find its speed in kmh-1, when it has moved through 100 m. Solution: Initial velocity = Vi = 40 ms -1Ģ.6 A train starts from rest with an acceleration of 0.5 ms -2. It comes to rest in the next 10 s with uniform deceleration. Total time = time to reach maximum height + time to return to the groundĢ.5 A car moves with a uniform velocity of 40 ms -1 for 5 s. (I) Maximum height attained by the ball S =? Solution: Initial velocity = V i = 30 ms -1Īcceleration due to gravity g = -10 ms -2 How long it will take to return to the ground? (45 m, 6 s) Calculate the maximum height reached by the ball. Solution: Initial velocity = V i = 10 ms -1Ģ.4 A tennis ball is hit vertically upward with a velocity of 30 ms -1, it takes 3 s to reach the highest point. Find the distance travelled during this time and the final velocity of the car. It accelerates at 0.2 ms -2 for half minute. What will be its speed at the end of 100 s.? (20 ms -1)Ģ.3 A car has a velocity of 10 ms -1. It moves through 1 km in 100 s with uniform acceleration. (80 s)Ģ.1 A train moves with a uniform velocity of 36 kmh-1 for 10 s. 2.10 In the above problem, find the time taken by the train to stop after the application of brakes.2.9 When brakes are applied, the speed of a train decreases from 96 kmh-1 to 48 kmh-1 in 800 m.(ii) initial velocity of the ball (45m, 30 ms-1).2.8 A cricket ball is hit vertically upwards and returns to ground 6 s later.Find the total distance travelled by train. 2.7 A train starting from rest, accelerates uniformly and attains a velocity of 48 kmh-1 in 2 minutes.2.6 A train starts from rest with an acceleration of 0.5 ms-2.2.5 A car moves with a uniform velocity of 40 ms-1 for 5 s.2.4 A tennis ball is hit vertically upward with a velocity of 30 ms-1, it takes 3 s to reach the highest point.It accelerates at 0.2 ms-2 for half minute. What will be its speed at the end of 100 s.? (20 ms-1) 2.1 A train moves with a uniform velocity of 36 kmh-1 for 10 s.
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