Please provide a clear explanation leted leted eted Class Management Help Februa

Question

Please provide a clear explanation

leted leted eted Class Management Help February 3 Begin Date: 2/1/2017 12:00:00 AM Due Date: 2/3/2017 2:00:00 AM End Date: 2/3/2017 2:00:00 AM (20%) Problem 5: A2.9-kg fireworks shell is fired at an angle 9.5° from the vertical from a mortar and reaches a height of 116 m. Take the upwards direction to be positive for this problem. Randomized Variables m 2.9kg 0.65 m a 9.5 33% Part (a) Neglecting air resistance (a poor assumption, but we will make it for this example) calculate the magnitude the shell's velocity when it leaves the mortar in m/s. Grade Summary Deductions Potential 100% Submissions Attempts remaining: S 4 5 6 cotano asin acos() per attempt) detailed view cosh() tanho cotanh0 Degrees O Radians I give up! deduction per feedback. Hints: deduction per hint Hints remaining: Feedback m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the 33% Part b) The self is a tube 0.65 mortar velocity found in part (a). il 33% Part (c) What is the average force on the shell in the mortar? Express your answer in newtons.

Explanation / Answer

a)

let's assume the initial velocity of launch = Vo

initial velocity in vertical direction is given as

Voy = Vo Cos9.5

a = acceleration = - 9.8

Y = displacement = 116 m

Vfy = final velocity = 0 m/s

using the equation

Vfy2 = Voy2 + 2 a Y

02 = (Vo Cos9.5)2 + 2 (-9.8) (116)

Vo = 48.35 m/s

b)

inside the mortar

Vi = initial velocity = 0 m/s

Vf = final velocity = 48.35 m/s

d = distance travelled = length of the tube = 0.65 m

a = acceleration

Using the equation

Vf2 = Vi2 + 2 a d

48.352 = 02 + 2 a (0.65)

a = 1798.25 m/s2

b)

average force is given as

F = ma

F = 2.9 x 1798.25 = 5214.925 N

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