Ask question

Ask question

All categories

2 years ago

11

A block of 250-mm length and 48 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b

ronze for which E = 95 GPa. Determine the largest load that can be applied, knowing that the normal stress must not exceed 80 MPa and that the decrease in length of the block should be at most 0.12% of its original length.

1 answer:

Marrrta [24]2 years ago

8 0

Answer:

153.6 kN

Explanation:

The elastic constant k of the block is

k = E * A/l

k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m

0.12% of the original length is:

0.0012 * 0.25 m = 0.0003 m

Hooke's law:

F = x * k

Where x is the change in length

F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)

The compressive load will generate a stress of

σ = F / A

F = σ * A

F = 80*10^6 * 0.048 * 0.04 = 153.6 kN

The smallest admisible load is 153.6 kN

You might be interested in

Why do areas north of the artic circle in tje northern hemisphere experience a polar day lasting for several months during summe

adell [148]

<span>The regioin is titled towqrd the Sun during polar day. (C)

(The same exact thing happens in areas south of the Antarctic Circle
in the southern hemisphere. The only difference is that the whole thing
is spelled better in the South.)</span>

7 0

2 years ago

Read 2 more answers

Can somebody please explain how a compass can be used to find North?

Ira Lisetskai [31]

if the pointy thingy in your compass is pointing north, that means it's being (pulled toward) something near Earth's north pole

5 0

2 years ago

You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh

Alexandra [31]

Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

$\beta =10\ log\dfrac{I}{10^{-12}}$

$\beta _1=10\ log\dfrac{I_1}{10^{-12}}$

$20=10\ log\dfrac{I_1}{10^{-12}}$

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

$\beta _2=10\ log\dfrac{I_2}{10^{-12}}$

$60=10\ log\dfrac{I_1}{10^{-12}}$

10⁶ x 10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity ,A=Area

$\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}$

$\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}$

r₂=0.1 m

4 0

2 years ago

Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on

Zolol [24]

Answer:71 dB

Explanation:

Given

sound Level $\beta _1=124 dB$

distance $r_1=5.01 m$

From sound Intensity

$\beta =10dB\log (\frac{I_1}{I_0})$

$124=10dB\log (\frac{I_1}{I_0})$

$12.4=\log (\frac{I_1}{I_0})$

$I_1=(1\times 10^{-12})\times 10^{12.4}$

$I_1=2.51 W/m^2$

we know Intensity $I\propto ^\frac{1}{r^2}$

$I_1r_1^2=I_2r_2^2$

$I_2=I_1(\frac{r_1}{r_2})^2$

$I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2$

$I_2=1.24\times 10^{-5} W/m^2$

Sound level corresponding to $I_2$

$\beta _2=10\log (\frac{I_2}{I_0})$

$\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})$

$\beta _2=70.93\approx 71 dB$

6 0

2 years ago

PLEASE HELP 100 POINTS

tia_tia [17]

Answer:

okay, So Um,

Explanation:

BTW, There are only 50 points, brainly subtracts half of the points

8 0

2 years ago