11. A12-kg object moving with of 6.0 m/s in the opposite aspeed of 8.0 m/s colli

Question

11. A12-kg object moving with of 6.0 m/s in the opposite aspeed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed the average force on the object by the wall? e direction. If the object is in contact with the wall for 2.0 ms, what is the magnitude of a) 9.8 kN b) 8.4kN c) 7.7 kN 9.1 kN e) 1.2 kN 12. of the horizontal impulse the catapult receives from the rock is A catapult fires an 800-kg rock with an initial velocity of 100 m/s at a 40* angle to the ground. The magnitude a) 5.1 x 104Ns ? 6.1 x 104 N . s. c)) 8.0 x 104 N-5 d) 5.0 x 10S N-s e) 60 x 105 N. 13. Two particles (m, 0.20 kg, m 0.30 kg) are positioned at the ends of a 2.0-m long rod of negligible mass. What is the moment of inertia of this rigid body about an axis perpendicular to the rod and mass? through the center of ah 0.48 kg m2 b 0.50 kg m2 c) 1.2kg m2 d) 0.30 kg·m2 e) 0.70 kg m2 14, A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at 0. The wheel has a diameter of 20 cm. What is the magnitude of the total linear acceleration of a point on the outer edge of the wheel at t # 060 s? a) 0.25 m/s2 b) 0.50 m/s2 c) 0.14 m/s2 d) 0.34 m/s2 el) 0.20 m/s2 15. The angular speed of the hour hand of a clock, in rad/min, is a) (l/1800)? d)? e) 120rt Turn in with answer sheet

Explanation / Answer

We know that

impulse = Favg*dt = change in momentum

Favg*dt = m*(v-Vo)

Given that

dt = 2*10^-3 sec

m = 1.2 kg

V = 8 m/s

Vo = -6 m/s

then

Favg*2*10^-3 = 1.2*(8+6)

Favg = 8400 N = 8.4kN

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