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e-lub [12.9K]
1 year ago
8

A 5.0-kg object is suspended by a string from the ceiling of an elevator that is accelerating downward at a rate of 2.6 m/s 2. W

hat is the tension in the string
Physics
1 answer:
Tcecarenko [31]1 year ago
8 0

The tension in the string is 36 N.

To find the answer, we need to know about pseudo acceleration.

<h3>What's pseudo acceleration?</h3>
  • When an object is placed on a vehicle, it experiences a pseudo acceleration in the opposite direction of the motion of the vehicle.
  • Magnitude of the pseudo acceleration is same as that of the acceleration of vehicle.
<h3>What's the pseudo acceleration of a mass suspended in a lift that's moving downward with an acceleration of 2.6 m/s²?</h3>
  • Here, the acceleration of the lift is 2.6m/s², so the mass suspended by a string inside it will feel a pseudo acceleration of 2.6m/s² upward.
  • So the effective acceleration of the mass = acceleration due to gravity - pseudo acceleration = 9.8 - 2.6 = 7.2 m/s²
<h3>What's the tension of the string suspending the mass?</h3>

Tension of the string= mass × effective acceleration

= 5 × 7.2 = 36 N

Thus, we can conclude that the tension in the string is 36 N.

Learn more about the pseudo acceleration here:

brainly.com/question/26644944

#SPJ4

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Three beads are placed along a thin rod. The first bead, of mass m1 = 28 g, is placed a distance d1 = 1.5 cm from the left end o
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Answer:

Part a)

Center of mass with respect to the left end is given as

r_{cm} = 5.63 cm

Part b)

Center of mass with respect to middle bead is

r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}

Part c)

Center of mass with respect to middle bead is

r_{cm} = 1.63 cm

Explanation:

Part a)

As we know that the center of mass of the system of mass is given by the formula

r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}

here we have

m_1 = 28 g

m_2 = 11 g

m_3 = 45 g

r_1 = 1.5 cm

r_2 = 1.5 + 2.5 = 4 cm

r_3 = 1.5 + 2.5 + 4.6 = 8.6 cm

Now we have

r_{cm} = \frac{28(1.5) + 11(4) + 45(8.6)}{28 + 11 + 45}

r_{cm} = 5.63 cm

Part b)

As we know that the center of mass of the system of mass is given by the formula

r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}

here we have

m_1 = 28 g

m_2 = 11 g

m_3 = 45 g

r_1 = -d_2 = -2.5cm

r_2 = 0

r_3 = d_3 = 4.6 cm

r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}

Part c)

Now plug in the values in above formula

r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}

r_{cm} = \frac{28(-2.5) + m_2(0) + 45(4.6)}{28 + 11 + 45}

r_{cm} = 1.63 cm

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Answer:

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Explanation:

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Effective resistance

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We have equation V = IR

Substituting

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