Ask question

Ask question

All categories

3 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?

1 answer:

LenKa [72]3 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

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

You might be interested in

Please help with this diagram as soon as possible.

melomori [17]

This is the answer .....

3 0

2 years ago

How does the tide cycle affect erosion along a sea coast?

Readme [11.4K]

<span>The tides are a constant pulling and pushing against the land. Causing erosion.</span>

4 0

3 years ago

The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p

Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

$4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$3.081x10^{16}m$

(c) How many meters in a light-year?

$9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$63325AU$

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

Since p is small it can be represent as:

$p(rad) = \frac{1AU}{d}$ (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

$p(rad) = \frac{p('')}{2.06x10^5}$ (2)

Then, equation 2 can be replace in equation 1:

$\frac{p('')}{2.06x10^5} = \frac{1AU}{d}$

$\frac{d}{1AU} = \frac{2.06x10^5}{p('')}$ (3)

From equation 3 it can be see that $1pc = 2.06x10^5 AU$

<em>a) How many parsecs are there in one astronomical unit? </em>

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

<em>(b) How many meters are in a parsec? </em>

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

<em>(c) How many meters in a light-year? </em>

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

$x = (3x10^{8}m/s)(31536000s)$

$x = 9.46x10^{15}m$

<em>(d) How many astronomical units in a light-year?</em>

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

<em>(e) How many light-years in a parsec?</em>

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

5 0

4 years ago

In the diagram, what is happening to the temperature at Point B? Question 6 options: A. The temperature is rising as the molecul

raketka [301]

<span> B. The temperature is not rising because the heat is being used to break the connections between the molecules </span>

6 0

3 years ago

Read 2 more answers

A stone is dropped from a cliff and falls 9.44 meters. What is the speed of the stone when it reaches the ground?

Lunna [17]

Answer:

option A

Explanation:

given,

height of the drop of stone = 9.44 m

speed of the stone = ?

As the stone is dropped the energy of the stone will be conserved.

using conservation of energy.

Potential energy = Kinetic energy

$m g h = \dfrac{1}{2} m v^2$

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8 \times 9.44}$

$v = \sqrt{185.024}$

v = 13.60 m/s

Hence, the correct answer is option A

3 0

3 years ago