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lina2011 [118]
2 years ago
10

A 3.50-m-long, 380kg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new b

uilding under construction. A 74.0kg construction worker stands at the far end of the beam.
What is the magnitude of the total torque due to the worker and the force of gravity on the beam about the point where the beam is bolted into place?
Physics
1 answer:
LenKa [72]2 years ago
8 0

Answer:

9064.44 Nm

Explanation:

Since the steel is uniform, the center of mass is also at the geometric center, which is 3.5 / 2 = 1.75 m from the bolted point.

Let g = 9.81 m/s2. The weight of the beam and the worker is

W_b = m_bg = 380*9.81 = 3727.8 N

W_w = m_wg = 74*9.81 = 725.94N

So the total torque about the bolted point is the sum of product of weights and distances from weight to the bolted point

W_b*1.75 + W_w*2.4 = 3727.8 *1.75 + 725.94 * 3.5 = 9064.44 Nm

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Answer:

The ball stops instantaneously at the topmost point of the motion.

Explanation:

Assume we have thrown a ball up in the air. For that we have given a force on the ball and it acquires an initial velocity in the upward direction.

The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.

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3 years ago
Four pairs of objects have the masses as described below, along with the distances between
lord [1]

Answer:

<h2>Mass of 1 Kg and 2 Kg, 1 meters apart.</h2>

Explanation:

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F=G\frac{m_{1} m_{2} }{r^{2} }

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Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min. Work out the difference be
Serggg [28]

Explanation:

We have,

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min.

1 mile = 1.6 km

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74 miles = 119.0 km

1 hour 15 min means 1.25 hours

Average speed of Ajoba is :

v_1=\dfrac{72.42 }{2.5}=28.96\ km/h

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v_2=\dfrac{119}{1.25}=95.2\ km/h

Difference in average speed of Ajoba and Prav is :

v=v_2-v_1\\\\v=v_2-v_1\\\\v=95.2-28.96\\\\v=66.24\ km/h

So, the difference in average speed of Ajoba and Prav is 66.24 km/h.

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