All categories

2 years ago

Draw the free-body diagram of the beam which supports the 80-kg load and is supported by the

1 answer:

5 0

The free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A can be seen in the image attached below.

The first image shows the diagram of the beam and the second image shows the free-body diagram of the beam.

The resolution of forces in the system is well understood by the principle of equilibrium where a stationary body will remain balanced when subject to parallel forces provided that the total sum of the overall external forces is zero.

The free-body diagram is a graphical representation used to visualize the forces applied to an object.

The equilibrium of forces on the x-axis is:

$\mathbf{\sum F_x = 0}$

The equilibrium of forces on the y-axis is:

$\mathbf{\sum F_y = 0}$

The equilibrium condition at any point is:

$\mathbf{\sum M = 0}$

From the free body diagram attached in the second image below,

the horizontal reaction is located at point A as $\mathbf{ A_x}$
the vertical reaction is located at point A as $\mathbf{A_y}$
the tension = T
the weight = W

Therefore, we can conclude that the free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A can be seen in the image attached below.

Learn more about the free-body diagram here:

brainly.com/question/19345060?referrer=searchResults

You might be interested in

HELP PLS

Angelina_Jolie [31]

Answer:

The correct option is;

B) Metamorphic Rocks

Explanation:

Zoisite, which is also referred to saualpite, is a metamorphic rock which is a hydroxy sorosilicate mineral formed from other types of rocks such as sedimentary, metamorphic and ingenious rocks in the process of their metamorphism under the presence high temperatures and pressures and mineral fluids which are hot

Zoiste is named after Sigmund Zois by Abraham Gottlob Werner in 1805 when Sigmund Zois sent Abraham Gottlob Werner the mineral specimen from Saualpe in 1805

6 0

3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after

kotegsom [21]

This question is incomplete, the complete question is;

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after 100 s, the reaction is 40% complete, how long (total time in seconds) will it take the transformation to go to 95% completion

y = 1 - exp( -ktⁿ )

Answer: the time required for 95% transformation is 242.17 s

Explanation:

First, we calculate the value of k which is the dependent variable in Avrami equation

y = 1 - exp( -ktⁿ )

exp( -ktⁿ ) = 1 - y

-ktⁿ = In( 1 - y )

k = - In( 1 - y ) / tⁿ

now given that; n = 2, y = 40% = 0.40, and t = 100 s

we substitute

k = - In( 1 - 0.40 ) / 100²

k = - In(0.60) / 10000

k = 0.5108 / 10000

k = 0.00005108 ≈ 5.108 × 10⁻⁵

Now calculate the time required for 95% transformation

tⁿ = - In( 1 - y ) / k

t = [- In( 1 - y ) / k ]^1/n

n = 2, y = 95% = 0.95 and k = 5.108 × 10⁻⁵

we substitute our values

t = [- In( 1 - 0.95 ) / 5.108 × 10⁻⁵ ]^1/2

t = [2.9957 / 5.108 × 10⁻⁵]^1/2

t = [ 58647.22 ]^1/2

t = 242.17 s

Therefore the time required for 95% transformation is 242.17 s

8 0

3 years ago

Of the cost reduction strategies for workers' compensation mentioned in the required readings, which one do you think would work

Vesnalui [34]

In industries together with production, we want people to address the manufacturing of merchandise and the usage of heavy machinery.

<h3>What is the painting situation?</h3>

In such painting situations, people are at risk of injuries, and this prices the maximum for the company. So so that you can put into effect value discount is such conditions we want to have right coincidence cowl plans for the people and make sure all of the protection precautions are taken withinside the factory.

The people have to be properly educated on using protection measures and in case any injuries arise we have to have coverage claims in order that we not want to make investments extra cash and we also can offer protection and protection to the people.
This approach is excellent for this enterprise due to the fact regardless of what number of precautions we take people are uncovered to fitness risks and as a result having the right coverage insurance is a superb value discount strategy.