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- Somewhere DEEP BELOW THE EARTH's surface, at an UNKNOWN displacement from the Earth's center, a particle of mass m is dangled from a long string, length L; the particle oscillates along a small arc according to the differential equation d^2x/dt^2=-(pi^2/36)x. Here x refers to an angular displacement measured from the vertical and t refers to time. The particle's mass is given by m=3kg. The length of the string is given by L=5 meters. Whenever the particle arrives at a location of x=(pi/12) radians from the vertical, the particle has no instantaneous speed. On both sides of the vertical, that is, x=(pi/12) radians is repeatedly observed to be a 'turning point' for the particle's periodic motion. 1. Draw a clear FREE-BODY diagram of this particle at some arbitrary point during oscillation, making sure to label variables and constants described above. 2. Approximating to three significant digits if necessary, what is the angular frequency of this oscillator on a string? 3. Approximating…A circular saw blade 0.6 m in diameter starts from rest and accelerates with constant acceleration to an angular velocity of 100 rad.s-1 in 20 s. Find the angle in degrees through which the blade has turned in the 20-s intervalApplying principles of dynamics (with step by steep solutions pls i give 5 stars) The displacement of a particle which moves along the x axis is given by x = (-4 + 3t2)e-0.25t, consider x to be in feet and t in seconds. Plot the displacement, velocity and acceleration for the first 20 seconds of motion and determine, both graphically and by your established equation for acceleration,the time at which acceleration is 0. (If the students find it more comfortable, they may choose to graph using excel or any graphing software, solutions however must be done manually)
- Graph below represents the force on a 12 kg air mass as it passes through a jet engine. Immediately after entering the engine, the force is 9,039 N, and the force tapers linearly to zero once the air reaches the back of the engine 2.3 m later. If the air enters the engine with velocity 252 m/s, what is its exit velocity? (Please answer to the fourth decimal place - i.e 14.3225)25-A wheel rotating with 11 rev/s is uniformly accelerated at 0.3 rad/s2 for 8 seconds. It then rotates with a uniform speed for next 11 s. Then it is made to stop in next 16 s. Determine (i)maximum angular velocity; (ii) total angular displacement Solution: The maximum velocty in rad/s is The total displacement in rad isFor the Following question Graph all 4 : [I just need all 4 graphs and please explain and make clean solution] Position vs time Velocity vs time Acceleration vs time Force vs time [For your convenience, I have solved the numerical solutions for the problem] (Please Look at the picture since it is much cleaner) Question : A 550 kilogram mass initially at rest acted upon by a force of F(t) = 50et Newtons. What are the acceleration, speed, and displacement of the mass at t = 4 second ? a =(50 e^t)/(550 ) [N/kg] v = ∫_0^t▒(50 e^t )dt/(550 )= v_0 +(50 e^t-50)/550=((e^t- 1))/11 x = ∫_0^t▒(e^t- 1)dt/(11 )= x_0 +(e^t- t - 1)/(11 ) a(4s)=(50*54.6)/550= 4.96[m/s^2 ] v(4s)=((e^4-1))/11= 4.87[m/s] x(4s)=((e^4- 4 - 1))/11= 4.51 [m]
- I have data accrued from a fan could you explain to me the concepts and show me how to calculate the horsepower and work done by the fan that data was composed of below in this table? Time (s) Run #1 Velocity, Blue (m/s) Run #1 0.05 0 0.1 0 0.15 0 0.2 0 0.25 0 0.3 0 0.35 4.04E-04 0.4 4.04E-04 0.45 0 0.5 -4.04E-04 0.55 -4.04E-04 0.6 0 0.65 0 0.7 4.04E-04 0.75 8.08E-04 0.8 4.04E-04 0.85 -4.04E-04 0.9 -8.08E-04 0.95 0 1 8.08E-04 1.05 4.04E-04 1.1 0 1.15 -4.04E-04 1.2 0 1.25 4.04E-04 1.3 0 1.35 0 1.4 -4.04E-04 1.45 0 1.5 4.04E-04 1.55 0 1.6 0 1.65 -4.04E-04 1.7 0 1.75 4.04E-04 1.8 0 1.85 0 1.9 -4.04E-04 1.95 0 2 4.04E-04 2.05 4.04E-04 2.1 4.04E-04 2.15 0 2.2 0 2.25 0 2.3 0 2.35 0 2.4 0 2.45 0 2.5 4.04E-04 2.55 8.08E-04 2.6 8.08E-04 2.65 8.08E-04 2.7 8.08E-04 2.75 8.08E-04 2.8 4.04E-04 2.85 0 2.9 0 2.95 0 3 0 3.05 0 3.1 0 3.15 0 3.2 0 3.25 0 3.3 0 3.35 0…A snowboarder’s velocity is tracked by a pulse-laser velocimeter as she descends the slope. These velocity data are stored and analyzed via an embedded curve fitting procedure to determine a velocity function of v(t) = 1.5t2 + 2t + 2 (t is time in seconds and v is velocity in ft/s). At time equal 0, position (x) is zero. Find her position (e.g., distance traveled), velocity and acceleration formulations and values of each when time equals 15 seconds. Clearly write the formulas before finding respective values at t = 15 seconds!The motion of a jet plane just after landing on a runway is described by the (v -t ) graph. Construct the (s —t ) and (a — t ) graphs for the motion. Here (s =0 ), when (t=0). =15t +ast v () V= ~15t+600 )
- A flywheel rotating at 500 rpm decelerates uniformly at a 3 rad/sec². How many seconds will it take for the flywheel to stop?* a. 49.36 s b. 49.35 s c. 49.37 s d. 49.40 sOne morning, Connie walks from her house to go to a store that is 500 ft, S 40o W from her house. She starts walking 120 ft, SW; then continues walking 100 ft, 10o S of E; then sees a friend who is 700 ft, 30o N of E from where she is. What is her friend’s location, (magnitude and direction) relative to her house? (Ans. 667.54 ft, 21.79o N of E) From her friend’s location, how far and in what direction should Connie walk to finally get to the store? (Ans. 1133.08 ft, 33.83o S of W)1. Circle weather the following statements are true or false. If the statement is true justify your reasoning if the statement is false correct the statement to make it true or justify why it is false a. Impulse is the integral of a force vs distance graph b. In two dimensions analyzing linear and angular momentum provides up to three scalar equations. c. Linear acceleration and angular acceleration both have units of ft/s^2 or m/s^2 d. When absolute motion or a rigid body it is important to find a function that relates linear and angular position then integrate to find velocity e. Instantaneous centers can only be determined for velocity and only for a snapshot in time