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3 years ago

6

Five identical springs, each with stiffness 420 N/m, are connected in parallel (that is, side by side) to hold up a heavy weight

. If these springs were replaced by an equivalent single spring, what should be the stiffness of this single spring?

1 answer:

balu736 [363]3 years ago

7 0

Answer:

Not surr

Explanation:

Idk

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When a severe weather watch has been issued by the National Weather Service, what should you do?

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4 years ago

Two mating steel spur gears are 20 mm wide, and the tooth profiles have radii of curvature at the line of contact of 10 and 15 m

Rom4ik [11]

Answer:

a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm

b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm

Explanation:

Given data:

L = 20 mm

F = 250 N

r₁ = 10 mm

r₂ = 15 mm

v = 0.3

E = 2.07x10⁵ MPa

$A=\frac{1-V_{1}^{2} }{E_{1} }-\frac{1-V_{2}^{2} }{E_{2} } =\frac{1-0.3^{2} }{2.07x10^{5} } *2=8.79x10^{-6}$

a) The maximum contact pressure is:

$P=0.564*\sqrt{\frac{F*(\frac{1}{r_{1} }+\frac{1}{r_{2} }) }{LA} } =0.564*\sqrt{\frac{250*(\frac{1}{10} +\frac{1}{15} )}{20*8.79x10^{-6} } } =274.58MPa$

The width of contact is:

$b=1.13*\sqrt{\frac{FA}{L(\frac{1}{r_{1} }+\frac{1}{r_{2} }) } } =1.13*\sqrt{\frac{250*8.79x10^{-6} }{20*(\frac{1}{10} +\frac{1}{15} )} } =0.029mm\\2*b=0.058mm$

b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is

T = 0.3*P = 0.3 * 274.58 = 82.37 MPa

At a distance of

0.8*b = 0.8*0.029 = 0.023 mm

6 0

3 years ago

The pressure in a compressed air storage tank is 1200 kPa. What is the tank's pressure in (a) kN and m units, (b) kg, m, and s u

Sever21 [200]

Answer:

a) 1200 kN/m²

b) 1,200,000 kg/ms²

c) 1.2 × 10⁹ kg/km.s²

Explanation:

Given:

Pressure = 1200 kPa

a) 1 Pa = 1 N/m²

thus,

1000 N = 1 kN

1200 kPa = 1200 kN/m²

b) 1 Pa = 1 N/m² = 1 kg/ms²

Thus,

1200 kPa = 1200000 Pa

or

1200000 Pa = 1200000 × 1 kg/ms²

or

= 1,200,000 kg/ms²

c) 1 km = 1000 m

or

1 m = 0.001 Km

thus,

1,200,000 kg/ms² = $\frac{1,200,000}{0.001}\frac{kg}{km.s^2}$

or

= 1.2 × 10⁹ kg/km.s²

3 0

3 years ago