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Lesechka [4]
3 years ago
6

The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t

he uniform distributed load that can be applied to the beam without risk of causing the strut to buckle. take f.s. = 2 against bucklin

Physics
1 answer:
EastWind [94]3 years ago
3 0

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

Ay = By = \frac{w * 6}{2} = 3w

P_c_r = 3w * F.S = 3w * 2.0 = 6w

I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7

To find the maximum intensity, w, let's take the Pcr formula, we have:

P_c_r = \frac{\pi^2 E I}{(KL)^2}

Let's take k = 1

E = 200*10^9

Substituting figures, we have:

6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}

Solving for w, we have:

w = \frac{67266.84}{6}

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

\sigma _c_r = \frac{w}{A}

\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y. This means it is safe

The maximum intensity w = 11.211KN/m

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While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction b
Citrus2011 [14]

Answer:

\theta = 66.7 degree

Explanation:

since force is applied downwards at some angle with the horizontal

so here we will have

F_n = mg + Fsin\theta

now we know that the box will not move if applied force is balanced by frictional force on it

so we will have

Fcos\theta = \mu F_n

F cos\theta = \mu (mg + F sin\theta)

F(cos\theta - \mu sin\theta) = \mu mg

F = \frac{\mu mg}{cos\theta - \mu sin\theta}

so here we can say

cos\theta - \mu sin\theta > 0

tan\theta = \frac{1}{\mu}

\theta = tan^{-1}\frac{1}{\mu}

\theta = tan^{-1}(\frac{1}{0.43})

\theta = 66.7 degree

6 0
3 years ago
A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in
Troyanec [42]

Answer:

The value is  \epsilon =  3.84 *10^{-5} \  V

Explanation:

From the question we are told that

  The diameter of the ring is  d =  18 \ mm  =  0.018 \  m

   The length of the solenoid is l = 25 \ cm  =  0.25 \ m

   The diameter of the solenoid is  D = 5.0 \ cm  = 0.05 \ m

    The number of turns is  N = 1500

   The change in  current in the solenoid is   \Delta  I   = 20 \ A

   The time taken is  \Delta  t  = 1 \ s

Generally the radius of the ring is  

     r = \frac{d}{2}

=>  r = \frac{0.018 }{2}

=>  r = 0.009 \ m

Generally the area of the ring is mathematically represented as  

      A = \pi r^2

=>   A = 3.142 *  0.009^2    

=>   A = 2.545 *10^{-4}\ m^2

Generally the induced emf is mathematically represented as

       \epsilon  =  A * \frac{dB}{dt}

Here    

         \frac{dB }{dt} =  \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}

Here  \mu_o is the permeability of free space with value  

         \mu_o =  4\pi *10^{-7} \ N/A^2

So  

     \frac{dB }{dt} =   4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}

=>  \frac{dB }{dt} =   0.150816\  T/s

So

     \epsilon =   0.150816 *  2.545 *10^{-4}

=>   \epsilon =  3.84 *10^{-5} \  V

3 0
3 years ago
Refrigerant-134a enters the expansion valve of a refrigeration system at 160 psia as a saturated liquid and leaves at 30 psia. D
KatRina [158]

Answer:

Temperature : 92.9 F

Internal Energy change: -2.53 Btu/lbm

Explanation:

As

mh1=mh2

h1=h2

In table A-11 through 13E

p2=120Psi, h1= 41.79 Btu/lbm,

u1=41.49

So T1=90.49 F

P2=20Psi

h2=h1= 41.79 Btu/lbm

T2= -2.43F

u2= 38.96 Btu/lbm

T2-T1 = 92.9 F

u2-u1 = -2.53 Btu/lbm

3 0
3 years ago
A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the
Shalnov [3]

Heat required to raise the temperature of a given system is

Q = ms\Delta T

here we know that

m = mass

s = specific heat capacity

\Delta T = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2

Q = 5 * 1700 * 25 + 2 * 900 * 25

Q = 212500 + 45000

Q = 257500 J

So heat required to raise the temperature of the system is 257500 J

4 0
3 years ago
Which term describes an observable fact, event, or circumstance?
madam [21]

Answer:

Phenomenon

Explanation:

phe·nom·e·non

/fəˈnäməˌnän/

noun

1.

a fact or situation that is observed to exist or happen, especially one whose cause or explanation is in question:

"glaciers are unique and interesting natural phenomena"

synonyms

occurrence, event, happening, fact, situation, etc.

2.

a remarkable person, thing, or event:

"the band was a pop phenomenon just for their sales figures alone"

synonyms

marvel, sensation, wonder, prodigy, miracle, etc.

3 0
3 years ago
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