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iragen [17]
3 years ago
12

1. Springs____________ energy when compressed And _________energy when they rebound.

Engineering
1 answer:
densk [106]3 years ago
6 0

Answer:

Springs store energy when compressed and release energy when they rebound

Explanation:

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A pin-supported structure has unrestrained rotations at the support locations.a) True b) False
antoniya [11.8K]

Answer:

a)True

Explanation:

Yes it is true a pin support can not resist the rotation motion . It can resist only lateral or we can say that only linear motion of structure and can not resit angular moment of motion about hinge or pin joint.On the other hand a fixed support can resist linear as well rotation motion of structure.

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3 years ago
Show What You Know: Creativity
tigry1 [53]

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3 years ago
A renewable item is something that is capable of being replaced naturally.
djverab [1.8K]
The answer is False.
6 0
4 years ago
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The viscosity of all liquids decreases as the temperature is a) Increased b) decreased c) maintained constant d) fluctuating e)
mars1129 [50]

Answer:

b) decreased

Explanation:

Viscosity is the measure of the resistance between the layers of the liquid. On increasing temperature, the kinetic energy of the molecules of the liquid increases which in turn increase in the random motion of the molecules. Thus,  the relative motion of  the molecules in the liquid become easier and hence, viscosity decreases.

5 0
3 years ago
If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical
lubasha [3.4K]

Answer:critical stress= 20.23 MPa

Explanation:

Since there was an internal crack, we will divide the length of the internal crack by 2

Length of internal crack, a = 0.7mm,

Half length = 0.7mm/2= 0.35mm  changing to meters becomes

0.35/ 1000= 0.35 x 10 ^-3m

The formulae for critical stress is calculated using

σC = (2Eγs /πa) ¹/₂

σC = critical stress=?

Given

E= Modulus of Elasticity= 225GPa =225 x 10 ^ 9 N/m²

γs= Specific surface energy = 1.0 J/m2 = 1.0 N/m

a= Half Length of crack=0.35 x 10 ^-3m

σC= (2 x 225 x 10 ^ 9 N/m² x 1.0 N/m /π x 0.35 x 10 ^-3m)¹/₂

=(4.5 x 10^11/π x 0.35 x 10 ^-3)¹/₂

=(4.0920 x10 ^14)¹/₂

σC=20.23 x10^6 N/m² = 20.23 MPa

​  

​

8 0
3 years ago
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