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Talja [164]
2 years ago
10

The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,

and AD = 15 lb. determine the force and moment acting at the trunk base point o. neglect the weight of the tree.

Physics
1 answer:
KATRIN_1 [288]2 years ago
3 0

Answer:

The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.

C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'

Explanation:

See attached picture.

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It has been proposed that extending a long conducting wire from a spacecraft (a "tether") could be used for a variety of applica
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The angle between shuttle's velocity and the Earth's field is  \theta =   24.2^o

Explanation:

From the question we are told that

     The length of eire let out is  L = 250 \ m

      The emf generated is \epsilon = 40 V

      The earth magnetic field is B = 5.0 *10^{-5} T

     The speed of the shuttle and tether is v =  7.80 * 10^3 \  m/s

The emf generated is mathematically represented as

                             \epsilon = L\ v\ B\ sin \ \theta

making \theta  the subject of the formula

                        \theta =   sin ^{-1}[ \frac{\epsilon}{L  * B  *v} ]

substituting values

                        \theta =   sin ^{-1}[ \frac{40}{250  * (5*10^{-5})  *(7.80 *10^{3})} ]

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When a car accelerates from a standing start, the crash test dummy appears to be pressed backward into the seat cushion. Which o
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<u>Answer:</u>

Option: D. Gravity is pulling the crash test dummy in the direction the car is moving.

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Answer:

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A stationary particle of charge q = 2.1 × 10-8 c is placed in a laser beam (an electromagnetic wave) whose intensity is 2.9 × 10
alisha [4.7K]
(a) The intensity of the electromagnetic wave is related to the amplitude of the electric field by
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(d) This time, the particle is moving with speed v=3.7 \cdot 10^4 m/s, in a direction perpendicular to the magnetic field (so, the angle \theta is 90^{\circ}), and so by using the intensity of the magnetic field we found in point (b), we can calculate the magnetic force on the particle:
F=qvB \sin \theta = (2.1 \cdot 10^{-8}C)(3.7 \cdot 10^4 m/s)(7.3 \cdot 10^{-3} T)(\sin 90^{\circ} )=
=5.7 \cdot 10^{-6} N
5 0
2 years ago
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