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TiliK225 [7]
4 years ago
9

According to the article "Edward R. Murrow: Inventing Broadcast Journalism," how did Murrow perceive the threat of Adolf Hitler?

Engineering
1 answer:
aliya0001 [1]4 years ago
8 0

Answer:

d. To Murrow, Hitler was a threat to all of civilization

Explanation:

For Murrow, Hitler's rise was a serious problem and a major threat to the entire civilization. For this reason, he believed that covering news about Hitler's advance and the battles of Nazi Germany was essential, even if the American population did not see Hitler as a threat and the European population, believed that these reprotations were only a way to denigrate the image of the continent to the world.

For Murrow reporting on Hitler's actions was as important as reporting on natural disasters.

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The specific volume of mercury is .00007 m^3/Kg. What is its density in lbm/ft^3?
aalyn [17]

Answer:

891.027 lbm/ft³

Explanation:

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5 0
3 years ago
Only an outer panel is being replaced. Technician A says that removing the spot welds by drilling through both panels allows the
Angelina_Jolie [31]

Answer:

6e66363636633747747363637737373737337374

5 0
3 years ago
A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o
victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width= 25 mm = 25\times 10^{-3} m

thickness = 6.5 mm = 6.5\times 10^{-3} m

crack length 2c = 0.5 mm at centre of specimen

\sigma _{applied} =  1000 N/cross sectional area

stress intensity factor  =  k  will be

\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}

                   = 6.154\times 10^{6} Pa

we know that

k =\sigma_{applied} (\sqrt{\pi C})

  =6.154\sqrt{\pi (2.5\times 10^{-04})}          [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

ifK_C = 1.15 Mpa m^{1/2} then load will be

Kc = \sigma _{frac}(\sqrt{\pi C})

1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}

\sigma _{frac} = 41.04 MPa

load = \sigma _{frac}\times Area

load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N

LAOD = 6669.86 N

3 0
3 years ago
a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8
djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam  along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we  determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = \frac{MC}{I}  =  \frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}  =  159.07 MPa

B) Determine max transverse shear stress

attached below

   ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam =  159.07 / 2 = 79.535 MPa  

attached below is the remaining solution

6 0
3 years ago
In DC electrode positive, how much power is at the work clamp?
Korolek [52]

Answer:

1/3 power

Explanation:

I'm just a smart guy

7 0
3 years ago
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