answer please! A small piece of Styrofoam packing material is dropped from aheig

Question

answer please!
A small piece of Styrofoam packing material is dropped from aheight of 1.80 m above the ground. Until it reaches terminal speed,the magnitude of its acceleration is given by a =g - bv. After falling 0.700 m, the styrofoameffectively reaches its terminal speed, and then takes 5.10 s moreto reach the ground. (a) What is the value of the constantb?
s-1

(b) What is the acceleration at t = 0?
m/s2 (down)

(c) What is the acceleration when the speed is 0.150 m/s?
3 m/s2 (down) answer please!
A small piece of Styrofoam packing material is dropped from aheight of 1.80 m above the ground. Until it reaches terminal speed,the magnitude of its acceleration is given by a =g - bv. After falling 0.700 m, the styrofoameffectively reaches its terminal speed, and then takes 5.10 s moreto reach the ground. (a) What is the value of the constantb?
s-1

(b) What is the acceleration at t = 0?
m/s2 (down)

(c) What is the acceleration when the speed is 0.150 m/s?
3 m/s2 (down)

Explanation / Answer

After falling thru 0.700 m, the onjet attains terminal speed,v (say). The remaining height, S = 1.80 - 0.70 = 1.10 m is covered in5.10 s at the uniform speed of v. Therefore, terminal speed,            v= 1.10/5.1 = 0.2157 m/s When the speed isterminal speed, acceleration,          a = g-b.v = 0 Hence,     b = g/v = 9.8/0.2157= 45.44/s (b) At t =0, initial velocity, u = 0 as the object isdropped.        a = g - b.v = g - b.u = g= 9.8m/s2 (c) At speed, v = 0.2157 m/s       a = g - b.v= 9.8 - 45.44x 0.15 = 9.8 - 6.816           =2.984m/s2 (c) At speed, v = 0.2157 m/s       a = g - b.v= 9.8 - 45.44x 0.15 = 9.8 - 6.816           =2.984m/s2

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