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3 years ago

If Fred has already measured 14 inches of a table measuring 30 inches. How many inches does he need to measure more?

Engineering

1 answer:

Olegator [25]3 years ago

8 0

If Fred already measured 14 inches and and the table is measuring 30 inches, Fred needs to measure 16 more inches

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How many sets of equations (V and M equations) would you need to describe shear and moment as functions of x for this beam? In o

den301095 [7]

Shear and moment as functions of x is described below .

Explanation:

1. Beam is the slender bar that carries transverse

loading; that is, the applied force are perpendicular to the bar.

2. In a beam, the internal force system consist of a shear force and

a bending moment acting on the cross section of the bar.

3. The shear force and the bending moment usually vary continuously

along the length of the beam.

4. The internal forces give rise to two kinds of stresses on a

transverse section of a beam:

(1) normal stress that is caused by

bending moment and

(2) shear stress due to the shear force.

Knowing the distribution of the shear force and the bending

moment in a beam is essential for the computation of stresses

and deformations.

Shear- Moment Equations

The determination of the internal force system acting at a given

section of a beam : draw a free-body diagram that expose these

forces and then compute the forces using equilibrium equations.

The goal of the beam analysis －determine the shear force V and the bending moment M at every cross section of the beam.

To derive the expressions for V and M in terms of the distance x

measured along the beam. By plotting these expressions to scale,

obtain the shear force and bending moment diagrams for the

beam.

The shear force and bending moment diagrams are convenient

visual references to the internal forces in a beam; in particular, they identify the maximum values of V and M

5 0

3 years ago

The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to

Sergeeva-Olga [200]

Answer:

hello below is missing piece of the complete question

minimum size = 0.3 cm

answer : 0.247 N/mm2

Explanation:

Given data :

section span : 10.9 and 13.4 cm

minimum load applied evenly to the top of span : 13 N

maximum load for each member ; 4.5 N

lets take each member to be 4.2 cm

Determine the max value of P before truss fails

Taking average value of section span ≈ 12 cm

Given minimum load distributed evenly on top of section span = 13 N

we will calculate the value of by applying this formula

= $\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12$ = 1.56 * 10^-5

next we will consider section ; 4.2 cm * 0.3 cm

hence Z (section modulus ) = BD^2 / 6

= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8

Finally the max value of P( stress ) before the truss fails

= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )

= 0.247 N/mm2

5 0

3 years ago

A transformer has 300,000 windings in its primary coil and uses 12,000V AC input. (4 points) How many windings would be needed t

viva [34]

Answer:

2750

Explanation:

The number of windings and the voltage are proportional.

Let n represent the number of windings to produce 110 Vac. Then the proportion is ...

n/110 = 300,000/12,000

n = 110(300/12) = 2750 . . . . multiply by 110

2750 windings would be needed to produce 110 Vac at the output.

7 0

2 years ago

Where can you find free air pods that look real

Tema [17]

You can find air pods that look real on letgo. or you can go to wish.com but if you want a good pair jus get the real ones

7 0

3 years ago

A 3.52 kg steel ball is tossed upward from a height of 6.93 meters above the floor with a vertical velocity of 2.99 m/s. What is

Dafna1 [17]

Answer : The final velocity of the ball is, 12.03 m/s

Explanation :

By the 3rd equation of motion,

$v^2-u^2=2as$

where,

s = distance covered by the object = 6.93 m

u = initial velocity = 2.99 m/s

v = final velocity = ?

a = acceleration = $9.8m/s^2$

Now put all the given values in the above equation, we get the final velocity of the ball.

$v^2-(2.99m/s)^2=2\times (9.8m/s^2)\times (6.93m)$

$v=12.03m/s$

Thus, the final velocity of the ball is, 12.03 m/s

7 0

3 years ago