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bekas [8.4K]
3 years ago
7

Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. De

termine the maximum shearing stress in shaft BC. (Round the final answer to one decimal place.) The maximum shearing stress in shaft BC is 56 MPa.

Physics
1 answer:
My name is Ann [436]3 years ago
3 0

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

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according to newton's second law of motion of the net force acting on the object increases while the mass of the object remains
Licemer1 [7]

Answer:

The Acceleration will increase

Explanation:

Newton's Second Law of motion: It states that the rate of change of momentum is directly proportional to the applied force and takes places along the direction of the force.

It can be expressed mathematically as,

F ∝ m(v-u)/t

Where (v-u)/t = a

F  = kma.

F = force, m = mass of the body, a = acceleration, k = constant of proportionality which tend to unity for a unit force, a unit mass, and a unit acceleration.

Therefore,

F = ma.

From the equation above,

If the net force acting on a body increase, while the mass of the body remains constant, the acceleration will also increase.

4 0
3 years ago
In an experiment the mass of a calorimeter is 36.35 g . Express in micrometer ,millimetre and kg.
Andru [333]

Answer:

1. 36.35 g = 36.35E15 micrometer.

II. 36.35 g = 363.5 millimetre.

III. 36.35 g = 0.03635 kilogram.

Explanation:

<u>Given the following data;</u>

  • Mass of calorimeter = 36.35 grams

To convert the mass in grams (g) to;

I. Micrometer

<u>Conversion:</u>

1 g = 1 exp 15 um

36.35 g = X um

Cross-multiplying, we have;

X = 36.35 * 1 exp 15 = 36.35 exp 15 um

<em>36.35 g = 36.35E15 micrometer</em>

II. Millimetre

<u>Conversion:</u>

1 g = 1 milliliter

36.35 g = X milliliter

Cross-multiplying, we have;

X = 36.35 * 1 = 36.35 milliliter

Next, we would convert milliliter to millimetre;

1 milliliter = 10 millimetre

36.35 milliliter = X millimetre

Cross-multiplying, we have;

X = 36.35 * 10 = 363.5 millimetre

<em>36.35 g = 363.5 millimetre</em>

III. Kilogram

<u>Conversion:</u>

1000 grams = 1 kilogram

36.35 g = X kilogram

Cross-multiplying, we have;

X * 1000 = 36.35 * 1

Dividing both sides by 1000, we have;

X = 36.35/1000 = 0.03635 kilogram

<em>36.35 g = 0.03635 kilogram</em>

<u>Note:</u>

  • g is the symbol for grams.
  • Exp (E) means exponential = 10
  • um is the symbol for micrometer.
4 0
3 years ago
Evidence suggests that there may be _______ momentum and ________ reversal patterns in stock price behavior.
bearhunter [10]

CORRECT ANSWER:

D. short-run; long run

STEP-BY-STEP EXPLANATION:

The whole question from book is

Evidence suggests that there may be _______ momentum and ________ reversal patterns in stock price behavior.

A. short-run; short-run

B. long-run; long-run

C. long-run; short-run

D. short-run; long run

Evidence suggests that there may be <em><u>short-run</u></em> momentum and <em><u>long run </u></em>reversal patterns in stock price behavior.

8 0
3 years ago
A fan blade, initially at rest, rotates with a constant acceleration of 0.029 rad/s2. What is the time interval required for it
LenKa [72]

Answer:

The time interval is  t = 21.30 \ s

Explanation:

From the question we are told that

    The constant acceleration is \alpha  = 0.029 \ rad / s^2

    The displacement is  \theta  =  6.58 \ rad

     

According to the second equation of motion we have that

    \theta  =  w_i* t  +  \frac{1}{2} *  \alpha  t^2

given that the blade started from rest

     w_i which is the initial angular velocity is 0

 So  

       \theta  = 0 +  \frac{1}{2} *  \alpha  t^2

 =>  t = \sqrt{ \frac{2 * \theta }{\alpha } }

substituting values  

=>    t = \sqrt{ \frac{2 * 6.58 }{0.029 } }

=>    t = 21.30 \ s

6 0
3 years ago
A 0.0600-kilogram ball traveling at 60.0 meters per second hits a concrete wall. What speed must a 0.0100-kilogram bullet have i
GarryVolchara [31]

Answer:

the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

Explanation:

The computation of the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is as follows:

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As there is a similar momentum

So,

\Delta Q = MV\\\\3.6 = 10^-2V\\\\V = 360 m/s

Hence, the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

5 0
2 years ago
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