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madreJ [45]
3 years ago
10

What else will change, if you change the point of view

Engineering
1 answer:
JulijaS [17]3 years ago
4 0

Answer:

We would need background context,

Explanation:

Then I would be happy to help!

You might be interested in
A metallic material with yield stress of 140 MPa and cross section of 300 mm x 100 mm, is subjected to a tensile force of 8.00 M
Readme [11.4K]

Answer:Yes,266.66 MPa

Explanation:

Given

Yield stress of material =140 MPa

Cross-section of 300\times 100 mm^2

Force(F)=8 MN

Therefore stress due to this Force(\sigma)

\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}

\sigma =266.66 \times 10^{6} Pa

\sigma =266.66 MPa

Since induced stress  is greater than Yield stress therefore Plastic deformation occurs

8 0
3 years ago
What is the angle of the input
olga55 [171]
Absolute positions — latitudes and longitudes
Relative positions — azimuths, bearings, and elevation angles
Spherical distances between point locations
3 0
3 years ago
Can i join three 12 volts batteriesto give me 24 volts output​
bulgar [2K]

Answer:

YES

Explanation:

If we connect batteries in series then the output voltage is the sum of the individual voltage of each battery i.e if you connect three 12 volts batteries in series then their output voltage will be 12+12+12=36 volts, but the current rating of the batteries in series will be same of the individual battery rating in 'mah'.

But when we connect the batteries in parallel their voltage is not added  but their current rating in mah is addition of their individual rating.

So, If you want 24 volts from three 12 volts battery then you can connect two of them in series and the other one in parallel with them this will give 24 volts and the current will be addition of the two series batteries and the third which is in parallel with them. You can use this configuration if current value is not a big factor.

8 0
3 years ago
Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify t
raketka [301]

Answer:

a) 53 MPa,  14.87 degree

b) 60.5 MPa  

Average shear = -7.5 MPa

Explanation:

Given

A = 45

B = -60

C = 30

a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

P1 = 53 MPa

Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

P1 = -68 MPa

Tan 2a = C/{(A-B)/2}

Tan 2a = 30/(45+60)/2

a = 14.87 degree

Principal stress

p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

b) Shear stress in plane

Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa

Average = (45-(-60))/2 = -7.5 MPa

5 0
3 years ago
The period of a pendulum T is assumed to depend only on the mass m, the length of the pendulum `, the acceleration due to gravit
zzz [600]

Answer:

The expression is shown in the explanation below:

Explanation:

Thinking process:

Let the time period of a simple pendulum be given by the expression:

T = \pi \sqrt{\frac{l}{g} }

Let the fundamental units be mass= M, time = t, length = L

Then the equation will be in the form

T = M^{a}l^{b}g^{c}

T = KM^{a}l^{b}g^{c}

where k is the constant of proportionality.

Now putting the dimensional formula:

T = KM^{a}L^{b}  [LT^{-} ^{2}]^{c}

M^{0}L^{0}T^{1} = KM^{a}L^{b+c}

Equating the powers gives:

a = 0

b + c = 0

2c = 1, c = -1/2

b = 1/2

so;

a = 0 , b = 1/2 , c = -1/2

Therefore:

T = KM^{0}l^{\frac{1}{2} } g^{\frac{1}{2} }

T = 2\pi \sqrt{\frac{l}{g} }

where k = 2\pi

8 0
3 years ago
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