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balandron [24]
3 years ago
5

For a bronze alloy, the stress at which plastic deformation begins is 294 MPa and the modulus of elasticity is 121 GPa. (a) What

is the maximum load that can be applied to a specimen having a cross-sectional area of 327 mm2 without plastic deformation? (b) If the original specimen length is 129 mm, what is the maximum length to which it may be stretched without causing plastic deformation?
Physics
1 answer:
FromTheMoon [43]3 years ago
7 0

Answer:

a) load in Newton is 96,138 b) 129.314mm

Explanation:

Stress = force/ area (cross sectional area of the bronze)

Force(load) = 294*10^6*327*10^-6 = 96138N

b) modulus e = stress/ strain

Strain = stress/ e = (294*10^6)/ (121*10^ 9) = 2.34* 10^ -3

Strain = change in length/ original length = DL/ 129

Change in length DL = 129 * 2.34*10^ -3 = 0.31347

Maximum length = change in length + original length = 129.314mm

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Titanium has a tensile strength of 63,000 PSI. ...

Chromium, on the Mohs scale for hardness, is the hardest metal around.

8 0
3 years ago
A bullet with a mass m b = 11.5 g is fired into a block of wood at velocity v b = 249 m/s. The block is attached to a spring tha
Ostrovityanka [42]

Answer:

0.358Kg

Explanation:

The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system

0.5Ke^2 = 0.5Mv^2

0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2

Using conservation of momentum between the bullet and the block

0.0115(265) = (M + 0.0115)v

3.0475 = (M + 0.0115)v

v = 3.0475/(M + 0.0115)

plugging into Energy equation

12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2

12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )

12.56 = 0.5 × 9.2872/ M + 0.0115

12.56 = 4.6436/ M + 0.0115

12.56 ( M + 0.0115 ) = 4.6436

12.56M + 0.1444 = 4.6436

12.56M = 4.6436 - 0.1444

12.56 M = 4.4992

M = 4.4992÷12.56

M = 0.358 Kg

4 0
3 years ago
A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude
miskamm [114]

Answer:

Explanation:

Given

Weight of person W=652\ N

At highest point Magnitude of the normal force F=585\ N

net force at highest point

F_{net}=W+F_c

where F_c= centripetal force

F_c=\dfrac{mv^2}{r}

F_{net}=F_{n}= Normal Force

585=652+F_c

F_c=-67\ N

Negative sign shows force is in upward direction

At bottom point centripetal force is towards the bottom

F_n=F_c+W

F_n=652+67

F_n=719\ N  

8 0
3 years ago
A satellite of mass 1000kg is in a circular orbit around a planet. The centripetal acceleration of the satellite in its orbit is
ludmilkaskok [199]

Answer: 5000N

Explanation:

The basic principle of a circular orbit is that Fg = m × ac, so as we have the mass and the centripetal acceleration (also called normal acceleration) we just have to operate. Fg = 1000kg × 5m/s² = 5000N

8 0
3 years ago
An object rolls East at a steady speed of 12 m/s for 3 seconds. What distance did the object travel
horrorfan [7]
Answer: 36 meters.


Equation to find distance:
Speed x time
8 0
3 years ago
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