3 Outlook ..ooo F 9:07 AM 78% moodle subredu [6] #1. Masses m, 195 kg and m, 584

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3 Outlook ..ooo F 9:07 AM 78% moodle subredu [6] #1. Masses m, 195 kg and m, 584 kg are located 68.5 cm apart. What is the magnitude of the force that m, exerts on m,? Answer a. 1.43 x 10' N. b. 87 x 10 N. c. 2.13 x 10 N. d. 2.39 x 10' N e, 2.97 x 10 N [5]#2. When two masses are a distance of96.0 m apart the force that one exerts on the other is 35 x 10 N. If the distance between the two masses suddenly increases to 3 times what it was (ie, the distance between them is now 288.0 m) but the mass of each is unchanged, what will be the new force between them? Answer a. 1.67 x 10' N. b. 2.783 x 10' N. c. 3.89x 10 N. d. 4.64 x 10 N. e, 5.71 x 10 N. [6] #3. Two charged particles Aand B are 250 nm apart. QA -4040e and Q 5840e. Choose the correct answer to complete the statement below. Statement: The dipole moment vector has magnitude C m and points from particle to particle Answer a. 1.41 x 10 24, A, B. b. 1.71 x 10 34, B, A. c. 2.35 x 10 A, B d. 2.69 x 10 34, A, B e. 3.55 x 1024, B, A. f For #4 and #5, use the figure at the right which represents two point charges Q +551C and Q -494 C, which are separated by a distance of 66.0 m. [6] #4. Determine the force on Q, due to Q. Answer: a 12 x 10 N, away from Q b. 2.82 x 10" N, away from Q, c, 4.19 x 10" N, toward Q e. 8.79 x 10 N, toward Q, f d. 6.16 x 10" N, toward Q. 18] #5. Determine the location where the net electric field due to Q,and Q, is zero? Answer: a. 15.0 m, right of Q b. 26.6 m, left of Q c. 42.3 m, left of Q d. 52.5 m eft of Q e. 73.2 m, right of Q, +68e The figure shows three situations (i), and (ii) involving [5] #6 +68e a charged spherical conducting shell with radius in terms of R, and a +68e

Explanation / Answer

(1).m1 = 195 kg

m2 = 584 kg

Distance between two masses r = 68.5 cm = 0.685 m

Required force F = Gm1m2/r 2

Where G = Gravitational constant = 6.67x10 -11 Nm 2/kg 2

Substitute values you get, F =(6.67x10 -11)(195)(584) /0.685 2

= 1.618x10 -5 N

(2). Force is inversely proportional to square of the distance between two masses.

So, F1/F2 = (r2/r1) 2

Given r2 = 3r1

F1/F2 = (3r1/r1) 2

= 9

F2 = F1/9

Given F1 = 35x10 -5 N

So, F2 = 3.88x10 -5 N

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