All categories

2 years ago

(15 Points)

Physics

1 answer:

oksian1 [2.3K]2 years ago

5 0

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

#SPJ1

You might be interested in

The graph below show how natural processes and human activities affect climate

Kaylis [27]

Is it just me or can anyone else not see a graph? (Because I can’t)

8 0

4 years ago

Read 2 more answers

A 3,600 kg sphere is located at a distance of 5.2 m from a 1,200 kg sphere.what is the gravitational force between the two spher

storchak [24]

Gravitational push or pull.

5 0

3 years ago

Based on the free-body diagram, the net force acting<br> on this firework is

Strike441 [17]

0N. The net force acting on this firework is 0.

The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.

Based on the free-body diagram, we have:

The force of gases is Fgases = 9,452N

The force of the rocket Frocket = -9452

Then, the net force acting is:

Fnet = Fgases + Frocket

Fnet = 9,452N - 9,452N = 0N

4 0

3 years ago

Read 2 more answers

Fill in the appropriate values for each blank as it refers to ATOM 1. The number of protons present in ATOM 1 is _________.

wlad13 [49]

3, protons are positive and there are 3 positive atoms visible

6 0

3 years ago

A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

In order to calculate the image position, we can use the lens equation:

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm$

The negative sign means that the image is virtual.

In order to calculate the image height, we use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

$y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm$

And the positive sign means the image is upright.

6 0

3 years ago