O study torque experimentally, you apply a force to a beam. One end of the beam is attached to a pivot that allows the beam to rotate freely. The pivot is taken as the origin or your coordinate system. You apply a force of F = Fx i + Fy j + Fz k at a point r = rx i + ry j + rz k on the beam. show answer No Attempt 33% Part (a) Enter a vector expression for the resulting torque, in terms of the unit vectors i, j, k and the components of F and r. τ = | γ θ i j k d Fx Fy Fz g m n rx ry rz ( ) 7 8 9 HOME ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + - 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 0% Potential 100% Submissions Attempts remaining: 3 (4% per attempt) detailed view Hints: 4% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback. No Attempt No Attempt 33% Part (b) Calculate the magnitude of the torque, in newton meters, when the components of the position and force vectors have the values rx = 0.76 m, ry = 0.035 m, rz = 0.015 m, Fx = 3.6 N, Fy = -2.8 N, Fz = 4.4 N. No Attempt No Attempt 33% Part (c) If the moment of inertia of the beam with respect to the pivot is I = 442 kg˙m2, calculate the magnitude of the angular acceleration of the beam about the pivot, in radians per second squared. All content © 2020 Expert TA, LLC

(a)  Resulting torque =  (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

(b) Magnitude of resulting torque  =  = 3.99 Nm

(c) angular acceleration = = 0.009027 rad/s²

Explanation:

Given Data;

I = 442 kg˙m2

rx = 0.76 m,

ry = 0.035 m,

rz = 0.015 m,

Fx = 3.6 N,

Fy = -2.8 N,

Fz = 4.4 N

F = Fx i + Fy j + Fz 1

r =  rx i + ry j + rz k 2

(a) Torgue is given by the formula;

T = r * F  3

Putting equation 1 and 2 into equation 3, we have;

Torque= r x F

            = (rx i +ry j +rz k) x (Fx i + Fy j +Fz k )

            = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

Therefore,

Resulting torque =  (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

b)

Putting given values into the above expression, we have

 torque =  (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

=(0.035*4.4 - (0.015*-2.8))i +(0.015*3.6 - 0.76*4.4)j+(0.76* -2.8 - 0.035*3.6)k

= (0.154 +0.041) i + (0.054 - 3.344) j + (-2.128 -0.126) k

= (0.196) i - (3.29) j + (-2.254) k

Magnitude of resulting torque = √(0.196² + 3.29² +2.254²

                                                  =√15.943031

                                                  = 3.99 Nm

c) Angular acceleration is given by the formula;

angular acceleration = torque/moment of inertia

                                   = 3.99/ 442

                                  = 0.009027 rad/s²

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2