Question

An object of mass m = 0.750 kg is suspended from the ceiling of an accelerating truck as in the figure below. Take a = 3.20 m/s. (a) Find the angle 8 (in degrees) that the string makes with the vertical. (b) Find the tension Tin N) in the string. What If? The truck returns to a constant velocity so that the mass hangs straight down inside the truck, whereby 9 = 0. Later, the truck carefully slows down. During this braking maneuver, the string is observed to make a constant angle of 9 = 23.0° with the vertical. (c) What is the acceleration of the truck (in m/s) while it is braking? m/s? (d) What is the direction of the deflection of the string while the truck is braking? forward backward Read It Master it 9. -/1 points SerPSE10 6.3.P.015.MI. A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 582 N. As the elevator later stops, the scale reading is 386 N. Assume the magnitude of the acceleration is the same during starting and stopping. (a) Determine the weight of the person. N (b) Determine the person's mass. (c) Determine the magnitude of acceleration of the elevator. m/s? Need Help? Read It Master It 10. -14 points SerPSE10 6.A.P.024. A car travels clockwise at constant speed around a circular section of a horizontal road as shown in the aerial view of the figure below. . (a) Find the directions of its velocity and acceleration at position direction of velocity -Select-- 1 direction of acceleration -Select-- 7 (b) Find the directions of its velocity and acceleration at position 6. direction of velocity --Select-- direction of acceleration --Select-- Need Help? Read It

Answer 1

. 19 C Given & mate of body, m= 0.750 kg. aceeleoa tion of toweli, az 3.20 m/s2 Past a) 6.8.0. q object from yoound frame. L . . along y-amit Ma Twoso = mg. – 0 ! Teos Tsino X-ARIS Tsino - ma - - Weight, Wing Dividing @ with © Toronto mg - » tano o% = 3.20 » [= 18.08) Answer from eq" O, Time = "*0006 0.750 x 9.3 7.73 N Past BD cepo NL Answer. 1. =) T = 7.73 Tension. part c) → while braking To Teasa a = 23° .: F.BD. 9 mess with respect to ground frame. - Long . O s & along yaxind, Twosa= mg along - 3 atsra. Как Tosina = ma. - © Bring Dividing a with @ ret mange :) tunaz gg =) a= gtan 23° => a = 4:16° Answer. Part (C) Direction of deflection of string in FORWARD part. d) Answer.

Questión - 9 when, ligt wstarts, weight measuoed, W, = 582 M. as elevator stope, weight measured, W, = 386N. Assuming, acceleration in both case = a. - masa of body, = m. . Case - when elevator starts Case - when elevator stops. 1 w W, - my ma mg - W₂ = ma l mg be £ 5. past a from Wi-mg = mg - tale » ang = W, + Wz » weight of body, and = mg = kl the = 'WE 489 N / Answer. mass of body, m= h 2 m2 49.38 kg Part b) me 49.32 kg 1 Anower Part - c) acceleration of body, a= Wi-mg = 582 -484 => [a: 1.98 m/s2) 49.33 (from eq -0). laz 1-98 m/3²/ Answer. Question - 10 part al Direction Direction of velocity = acceleration = East South Past B) Direction Direction z of velocity, acceleration South = Wenst.