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Ivenika [448]
3 years ago
9

Technician A says the counter gear is actually several gears machined out of a single piece of steel. Technician B says the coun

ter gear is driven by the clutch (main drive/input) gear and drives the mainshaft speed gears. Who is correct?
a. Technician A only
b. Technician B only
c. Both A and B
d. Neither A nor B
Physics
2 answers:
Lesechka [4]3 years ago
8 0

Answer:

The right option is C. Both A and B as the counter gear is a manual transmission gear with several gears on a rod to drive the mainshaft speed gears

Explanation:

The counter gear is a manual transmission shaft, with an opposite direction of rotation to that of the engine, consisting of several gears that are forged along a gear rod base and located at the bottom of the transmission system. It affords a connection from the transmission input shaft to the output shaft with control from the clutch as such it drives the output speed gears

dexar [7]3 years ago
4 0

Answer:

Option c. (Both Technician A and B are correct)

Explanation:

A transmission system consists of 3 shafts. The input shaft, the counter shaft, and the main shaft. The clutch gear always rotates with input shaft and is a crucial element of the input shaft.

The counter shaft is actually several gears machined out of a single piece of steel. The counter shaft may also be called counter gear or cluster gear.  It is a secondary shaft that runs parallel to the mainshaft in a gearbox and is used to provide powers to machine components such as the drive axle.

The main gears (also called the speed gears) on main shaft (also known as the output shaft) are used to transfer rotation from counter shaft to the output shaft.

Hence in the light of above description, both technician A and B are correct.

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A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym
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Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

                                     L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

                                    L_(p,i) = m*V_pi*R

                                    L_(p,i) = 3.6*3.3*0.44

                                    L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

                                    L_(c,i) = I*w_i

                                    L_(c,i) = 0.5*M*R^2*0

                                    L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

                                    L_i = L_(p,i) + L_(c,i)

                                    L_i = 5.2272 + 0

                                    L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

                                   L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

                                   L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

                                  L_f = L_(p,f) + L_(c,f)

                                  L_f = I_p*w_f + I_c*w_f

                                  L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

                                  L_i = L_f

                                  5.2272 = w_f*(I_p + I_c)

                                  w_f =  5.2272/ R^2*(m + 0.5M)

Plug in values:

                                  w_f =  5.2272/ 0.44^2*(3.6 + 0.5*45)

                                  w_f =  5.2272/ 5.05296

                                  w_f = 1.0345 rad/s

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2 years ago
If santa slides 5.00 m before reaching the edge, what is his speed as he leaves the roof?
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Even with no friction, it depends on the slope of the roof. That is, it depends on how much elevation (altitude) he loses during the slide.

Whatever that number is ... call it 'h' ... Santa's speed when he reaches the edge is

Square root of (19.6h) meters per second.

It doesn't matter how much he weighs, or how far he has slud. Only how much altitude he lost on the slope while sliding.

8 0
2 years ago
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