Collin College My Courses Question Completion Status OD 289

Collin College My Courses Question Completion Status: OD. -2.89 x 109 N.m QUESTION 13 0.5 points Save Ans A 2.8-N force is applied to the plunger of a hypodermic needle. If the diameter of the plunger is 1.3 cm and that of the needle 0.20 mm, with what force does the fluid leave the needle? OA. 2.37 x 10-4 N O B. 4.31 x 10-2 N OC. 6.63 N OD, 6.63 x 10-4 N 0.5 points QUESTION 14 A 0.650-kg mass oscillates according to the equationx- 0. 25sin(5.50t) where x is in meters and t is in seconds. Determine (a) the amplitude, (b) energy when x is 15 cm. the frequency, (c) the period, (d) the total energy, and (e) the kinetic energy and potential 0 A. (a) A-0.25 m, (b) f= 0.88 Hz, (c) T-1.14 s, (d) E-0.61 J, (e) U-0.39 J and K-0.22 B. (a) A-0.25 m, (b) f-0.88 Hz, (c) T = 1.14 s, d) E-0.61 J, (e) U-0.22 J and K 0.39 J O c. (a) A-5.00 m, (b) f= 0.88 Hz, (c) T-1.14 s, (d) E-0.61 J, (e) U-0.22) and K-0.39 J 0 D. (a) A = 0.25 m, (b) f= 1.14 Hz. (c) T-0.88 s, (d) E = 0.61 J, (e) u-0.22 J and K = 0.39 J QUESTION 15

Solution

13)

Pressure needs to same.

So, F1/A1 = F2/A2

So, 2.8/(pi*0.013^2/4) = F2/(pi*0.0002^2/4)

So, F2 = 6.63*10^-4 N

14)

Amplitude = 0.25 m

Frequency , f= 5.5/(2*pi)

= 0.875 Hz = 0.88 Hz

Time period ,T = 1/f = 1/0.875

= 1.14 s

Total energy = 0.5*0.65*(5.5*0.25)^2

= 0.614 J

Kinetic energy when x = 15 cm= 0.15 m

At this moment, t = asin(0.15/0.25) / 5.5 = 0.117 s

So, at this time velocity , V = 0.25*5.5*cos(5.5*0.117)= 1.1 m/s

So, kinetic energy = 0.5*0.65*1.1^2 = 0.39 J

And potential energy = 0.61 - 0.39 = 0.22 J

15)

There is new amplitude (that is 0.15 m) is 1.5 times the original(0.10m).

Now kinetic energy is prortional to square of amplitude.

So, new kinetic energy = 1.5^2 = 2.25 times

16)

100 mosquitoes will make 100 times the noise.

So, in dB, 10*log(100) = 20 dB