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klemol [59]
3 years ago
11

For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vec

tors? For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vectors? a) The velocity vector is perpendicular to the acceleration vector; the acceleration vector is perpendicular to the net force vector. b) The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector. c) The velocity vector is parallel to the acceleration vector; the acceleration vector is perpendicular to the net force vector. d) The velocity vector is parallel to the acceleration vector; the acceleration vector is parallel to the net force vector.
Physics
1 answer:
zavuch27 [327]3 years ago
3 0

Answer: b) The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector.

Explanation: A change in velocity creates an acceleration. As the object rotates through the circular path it is constantly changing direction, and hence accelerating, which causes a constant force to act upon the object. This Force acts towards the center of curvature, directly toward the axis of rotation in a direction parallel to the acceleration of the body along the path. Because the object is moving perpendicular to the force, the path followed by the object is a circular one. Hence the velocity of the object is perpendicular to the acceleration.

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A machine pulls a 46 kg trunk 3.0 m up a 42o ramp at constant velocity, with the machine's force on the trunk directed parallel
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Answer:

the answers are

W = 1271.256 J

E_{th}= 361.81 J

Explanation:

The mass of the trunk (m) = 46kg

angle between the ramp and the horizontal (θ) = 42°

The trunk is pulled at a displacement (d) up the ramp = 3m

The coefficient of kinetic friction between the trunk and the ramp = 0.36

The trunk is moving with a constant velocity therefore the net force on it is zero, therefore the force required to move the trunk must be equal to the summation of forces opposing the trunk

The two forces opposing the trunk are

  1. the gravitational force directed down the ramp  F_{g} and
  2. the frictional force between the ramp and the trunk F_{f}

We have to calculate the machine's force which is equal to sum of the

F =F_{g} + F_{f}

F_{g} = mgsinθ

F_{f} =μ× N

N = mgcosθ

F_{f} = μmgcosθ

F = mg (sinθ + μcosθ)

F = 46× 9.8 (sin42 + 0.36×cos42)

F= 450.8 (0.67 +0.27)

F = 450.8 × 0.94

F = 423.752N

to calculate the workdone on the trunk by the machine force

The workdone on the trunk is  = W = F × dcosΘ

Θ = 0° because the trunk is directed parallel to the ramp

W =423.752× 3 cos 0

W = 423.752 ×3

W = 1271.256 J

(b)  the increase in thermal energy of the trunk and the ramp

Friction converts mechanical energy into thermal energy, so multiplying the frictional force with the distance gives the thermal energy generated by the trunk

E_{th} =F_{f} × d

= μmgcosθ ×d

= 0.36 ×46×9.8×cos42×3

= 361.81 J

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Answer:

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Explanation:

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Answer:

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Explanation:

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Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass
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Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

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So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

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Answer:  Scattering reflection

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Explanation:

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