Answer:
Simple awnser Do it yourself I really would help but I have no clue! Sorry
Explanation:
The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
- Let's draw the free body diagram of the system using the given data.
- From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
- For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

- We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

- To find Ny, we need to find the tension T.
- For this, we can equate the net horizontal force.

- Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

- Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
Learn more about the tension here:
brainly.com/question/28106871
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Answer:
2000 nickels
Explanation:
One way to solve proportionality problems, direct and inverse: the simple 3 rule.
If the relationship between the magnitudes is direct (when one magnitude increases so does the other), the simple direct rule of three must be applied.
On the contrary, if the relationship between the magnitudes is inverse (when one magnitude increases the other decreases) the rule of three simple inverse applies.
The simple 3 rule is an operation that helps us quickly solve proportionality problems, both direct and inverse.
To make a simple rule of three we need 3 data: two magnitudes proportional to each other, and a third magnitude. From these, we will find out the fourth term of proportionality.
In the simple three rule, therefore, the proportionality relationship between two known values A and B is established, and knowing a third value C, a fourth value D is calculated.
A -> B
C -> D
Calculation
1 nickel --> 5 g
X? nickel --> 10000g
X = (10000 g * 1 nickel) / 5 g
X = 2000 nickels
vib. motion motion of wire of guitar
circular motion revolution of earth around sun
1ml 1cm3
1m3 100cm3
volume of liquid measuring cylinder
Answer:
Explanation:
Calculating the exit temperature for K = 1.4
The value of
is determined via the expression:

where ;
R = universal gas constant = 
k = constant = 1.4


The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :
------ equation(1)
we can rewrite the above equation as :


where:



Thus, the exit temperature = 402.36 K
The exit pressure is determined by using the relation:



Therefore, the exit pressure is 17.79 bar