s. A005 kg steel ball vibrating rapidly up and down with small amplisunle. The b

Question

s. A005 kg steel ball vibrating rapidly up and down with small amplisunle. The ball beariag wered to a height of 12 m above the plate. What was the upward irmpai bounces back up delivered to the ball bearing during the collision with the plate? )0.021N b) 022Ns c) 0.24 Ns d) 046 Ns e) insufficient information given er ne 9. A satellite eari lis vih notate-m2 e Initally panng at 12 rad year it is spinning at 5.8 rad/s. What is the average retarding tonj during the year? torque exerted on the craft a) -930 N-m b) -2.95 x 10 N-m c) -196x 107 N-m d) -0.04 N-m e) -24 N-m 10. A 3-meter long ladder of mass 8 kg leans against a wall. If the static friction coefficient between ladder and ground is 05 and the wall friction is negligible, what is the s angle of the ladder with respect to the ground for the ladder to remain in equilibrium? a) 27 b) 38 c) 45 d) 63° c) 90° 11. In the figure below,five forces act on a massless rod free to pivot at point P. Which force is producing a clockwise torque about point P? PO

Explanation / Answer

8. v = sqrt(2 g h )

vi = -sqrt(2 x 9.8 x 1) = -4.43 m/s

vf = sqrt(2x 9.8 x 1.2) = 4.85 m/s

Impulse = m (vf - vi) = 0.05 (4.85 - (-4.43))

= 0.46 N s

Ans(d)

9. wf = wi + alpha t

5.8 = 12 + alpha(1 x 365 x 24 x 3600 s)

alpha = - 1.966 x 10^-7 rad/s^2

torque = I alpha = - 2.95 x 10^-5 N m

Ans(b)

10. N = m g

and fs_max = u N = u m g = 0.5 x 8 x 9.8 = 39.2 N  

Nw = fs_max = 39.2 N

balancing moment about the floor,

M = (L/2 m g cos(theta)) - (L Nw sin(theta)) = 0

(8 x 9.8 / 2) cos(theta) = 39.2 sin(theta)

tan(theta) = 1

theta = 45 deg Ans(c)

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.