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4 years ago

5

Welding, cutting, or heating in any enclosed spaces involving which of the following metals should be performed with local exhau

st ventilation or employees should be provided air line respirators:

1 answer:

sweet [91]4 years ago

4 0

steal aluminum iron

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Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3

Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

$\tau=\sigma\cos\alpha\cos\beta$

Put the value into the formula

$\tau=12\cos60\times\cos35$

$\tau=4.91\ MPa$

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

$\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}$

Put the value into the formula

$\sigma=\dfrac{6.2}{\cos60\cos35}$

$\sigma=15.13\ MPa$

Hence, This is the required solution.

3 0

3 years ago

ich of the following is not accurate when describing solids? A. The amount of pressure exerted by a solid is solely dependent on

Dafna11 [192]

The amount of solid does not affect how you are describing the solid so a is the answer

5 0

3 years ago

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

8 0

3 years ago

12) Photosynthesis is a chemical reaction where carbon dioxide and water react to form glucose (C6H12O6) and oxygen gas. Which r

marusya05 [52]

6CO2 + 6H2O → C6H12O6 + 6O2

4 0

4 years ago

Read 2 more answers

Calculate the average induced voltage between the tips of the wings of an airplane flying above East Lansing at a speed of 885 k

Yakvenalex [24]

Answer:

=0.855V

Explanation:

The induced voltage can be calculated using below expression

E =B x dA/dt

Where dA/dt = area

B= magnetic field = 6.90×10-5 T.

We were given speed of 885 km/h but we will need to convert to m/s for consistency of unit

speed = 885 km/h

speed = 885 x 10^3 m/hr

speed = 885 x 10^3/60 x60 m/s

speed = 245.8 m/s

If The aircraft wing sweep out" an area

at t= 50.4seconds then we have;

dA/dt = 50.4 x 245.8

= 123388.32m^2/s

Then from the expression above

E =B x dA/dt substitute the values of each parameters, we have

E = 6.90 x 10^-5 x 12388.32 V

E =0.855V

Hence, the average induced voltage between the tips of the wings is =0.855V

6 0

3 years ago