please exam tomm. i am so confused...im not a science majorbut have to take phys

Question

please exam tomm. i am so confused...im not a science majorbut have to take physics.


A)A 520 N cat burglar is supported bycables in the figure below. Assume the angle ofthe inclined cable is 32.0°. (1) Find the tension in each cable. inclined cable 1 N horizontal cable 2 N vertical cable 3 N
(2) Suppose the horizontal cable were reattached higher up on thewall. Would the tension in the inclined cable increase, decrease,or stay the same?
4 increase
decrease    
stay the same



B)A shopper in a supermarket pushes a loaded cart with a horizontalforce of 9 N. The cart has a mass of25 kg. (1) How far will it move in 6.0 s, starting from rest? (Ignore friction.)
1 m
(2) How far will it move in 6.0 s if theshopper places his 30 N child in the cart before he begins to pushit?
2 m


C)A 8.0 kg object undergoes anacceleration of 2.9 m/s2. (1) What is the magnitude of the resultantforce acting on it?
1 N

(2) If this same force is applied to a 16.0 kg object, what acceleration is produced?
2 m/s2



4 (1) How far will it move in 6.0 s, starting from rest? (Ignore friction.)
1 m
(2) How far will it move in 6.0 s if theshopper places his 30 N child in the cart before he begins to pushit?
2 m


C)A 8.0 kg object undergoes anacceleration of 2.9 m/s2. (1) What is the magnitude of the resultantforce acting on it?
1 N

(2) If this same force is applied to a 16.0 kg object, what acceleration is produced?
2 m/s2



(1) What is the magnitude of the resultantforce acting on it?
1 N

(2) If this same force is applied to a 16.0 kg object, what acceleration is produced?
2 m/s2 inclined cable 1 N horizontal cable 2 N vertical cable 3 N

Explanation / Answer

Horizontal: T2 = T1 (cos180/cos32)= T1 (-1/0.848) = -1.179T1                   T1sin180 + (-1.179T1)(sin 32) - 520N = 0                   0+ (-1.179T1)(0.5299) - 520N                   0.625T1 =520N/0.625                   T1= 832 N
Incline:    T2 =1.179T1 = (1.179)(832N)                   =980.928 N Vertical: 520 N 2) decrease B) shopping cart question 1)ax = 9N/25kg = 0.36 m/s2 deltaX = 1/2axt2 =1/2(0.36)(6s)2       = 6.48 m 2) 30N/9.8 = 3.06 + 25 kg = 28.06 kg (new mass of cart) 9N/28.06kg = 0.321 m/s2 1/2(0.321)(6s)2 = 5.778 m C) acceleration question 8kg x 2.9 m/s2 = 23.2N 23.4N/16kg = 1.4625 m/s2 1/2(0.321)(6s)2 = 5.778 m C) acceleration question 8kg x 2.9 m/s2 = 23.2N 23.4N/16kg = 1.4625 m/s2

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