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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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The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th
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According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

E_i = E_f

0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

PART B) Replacing the values given as,

h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}

v_f = 1.8486m/s

Therefore the speed of the masses would be 1.8486m/s

6 0
3 years ago
describe at least two ways in which parents are able to influence the drinking habits of their children
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A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
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Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

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