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larisa86 [58]
3 years ago
8

The flexural strength or MOR of a ceramic is 310 MPa. A block of the ceramic, which is 20 mm wide, 15 mm high, and 300 mm long,

is supported between two rods 150 mm apart. Determine the force required to fracture the material, assuming no plastic deformation occurs.

Engineering
1 answer:
Alja [10]3 years ago
3 0

Answer:

F=6200\ \text{N}\\

Explanation:

In this problem you need to define the force that acts upon a beam in a 3 point bending problem. I put a picture of the problem taken from Wikipedia:

In this problem the flexural strength is defined with the following formula:

\sigma=\cfrac{3FL}{2bd^2}

where F is the force applied, L the length between the two rods, b the width of the ceramic block and d it's height.

The force is then defined as:

F=\cfrac{2\sigma bd^2}{3L}=6200\ \text{N}

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A crankcase heater is often used to prevent refrigerant from mixing with compressor oil during periods of:
stira [4]

Answer:

Low ambient temperature

Explanation:

Hope this helps. If it did, please mark as brianliest so other people see it. Thanks! - Kai

5 0
3 years ago
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A 3-ft-diameter vertical cylindrical tank open to the atmosphere contains 1-ft-high water. The tank is now rotated about the cen
arlik [135]

Answer:

The angular velocity is 7.56 rad/s

the maximum water height is 2 ft

Explanation:

The z-position as a function of r is equal to

z_{s(r)} =h_{0} -\frac{w^{2}(R^{2}-2r^{2}   }{4g} (eq. 1)

where

h0 = initial height = 1 ft

w = angular velocity

R = radius of the cylinder = 1.5 ft

zs(r) = 0 when the free surface is lowest at the centre

Replacing and clearing w

w=\sqrt{\frac{4gh_{0} }{R^{2} } } =\sqrt{\frac{4*32.17*1}{1.5^{2} } } =7.56rad/s

if you consider the equation 1 for the free surface at the edge is equal to

z_{s(R)} =h_{0} +\frac{w^{2}R^{2}   }{4g} =1+\frac{(7.56^{2})*(1.5^{2} ) }{4*32.17} =1.99ft=2ft

7 0
3 years ago
Consider a resistor made of pure silicon with a cross-sectional area pf 0.5 μm2, and a length of 50 μm. What is the resistance o
lukranit [14]

Answer: 24 pA

Explanation:

As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.

Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵  Ω  cm.

The resistance R of a given resistor, is expressed by the following formula:

R = ρ L / A

Replacing by the values for resistivity, L and A, we have

R = 2.1. 10⁵ Ω  cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2

R = 2.1. 10¹¹ Ω

Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:

I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA

7 0
3 years ago
It has a piece of 1045 steel with the following dimensions, length of 80 cm, width of 30 cm, and a height of 15 cm. In this piec
Serggg [28]

Answer:

material remove in 3 min is 16790.4 mm³/s

Explanation:

given data

length L = 80 cm = 800 mm

width W = 30 cm

height H = 15 cm

make grove length = 80 cm

width = 8 cm

depth = 10 cm

mill toll diameter = 4 mm

axial cutting depth = 20 mm

to find out

How much material removed in 3 minutes

solution

first we find time taken for length of advance that is

time = \frac{length}{advance}

here advance is given as 0.001166 mts / sec

so  time = \frac{800}}{0.001166*1000}

time = 686.106 seconds

now we find material remove rate that is

remove rate = mill toll rate × axial cutting depth × advance

remove rate = 4 × 20×0.001166 ×1000

remove rate = 93.28 mm³/s

so

material remove in 3 minute = 3 × 60 = 180 sec

so material remove in 3 min = 180 × 93.28

material remove in 3 min is 16790.4 mm³/s

7 0
3 years ago
True power can only be measured across what?
loris [4]

Answer: mets

Explanation: meets are good

6 0
3 years ago
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