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Ksenya-84 [330]
3 years ago
5

If the frequency of a system undergoing simple harmonic motion doubles, by what factor does the maximum value of acceleration ch

ange?
a. 4
b. 2/pi
c. 2
d. (2)^1/2
Physics
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

the answers the correct one is a  4

Explanation:

The centripetal acceleration is by

           a = v² / R

angular and linear velocities are related

           v = w R

let's substitute

           a = w² R

for initial condition

           a₀ = w₀² R

suppose the initial angular velocity is wo, suppose the angular velocity doubles

           a = (2w₀)² R

           a = 4 (w₀² R)

           a = 4 a₀

when reviewing the answers the correct one is a

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A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s r
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Answer:

There's a decrease in width of 2.18 × 10^(-6) m

Explanation:

We are given;

Shear Modulus;E = 207 GPa = 207 × 10^(9) N/m²

Force;F = 60000 N.

Poisson’s ratio; υ =0.30

We are told width is 20 mm and thickness 40 mm.

Thus;

Area = 20 × 10^(-3) × 40 × 10^(-3)

Area = 8 × 10^(-4) m²

Now formula for shear modulus is;

E = σ/ε_z

Where σ is stress given by the formula Force(F)/Area(A)

While ε_z is longitudinal strain.

Thus;

E = (F/A)/ε_z

ε_z = (F/A)/E

ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))

ε_z = 3.62 × 10^(-4)

Now, formula for lateral strain is;

ε_x = - υ × ε_z

ε_x = -0.3 × 3.62 × 10^(-4)

ε_x = -1.09 × 10^(-4)

Now, change in width is given by;

Δw = w_o × ε_x

Where w_o is initial width = 20 × 10^(-3) m

So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

Δw = -2.18 × 10^(-6) m

Negative means the width decreased.

So there's a decrease in width of 2.18 × 10^(-6) m

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Which causes the thermocline?
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Answer:

Random Motion is a motion in which an object didn't go in a straight manner, for ex: zig zag lines, curved, etc.

Explanation:

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A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness
Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

\sigma_h = \frac{Pd}{2t}

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

\sigma_l = \frac{Pd}{4t}

The letters have the same meaning as before.

Then he hoop stress would be,

\sigma_h = \frac{Pd}{2t}

\sigma_h = \frac{75 \times 8}{2\times 0.25}

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And the longitudinal stress would be

\sigma_l = \frac{Pd}{4t}

\sigma_l = \frac{75\times 8}{4\times 0.25}

\sigma_l = 600Psi

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

\tau_{max} = \frac{\sigma_h}{2}

\tau_{max} = \frac{1200}{2}

\tau_{max} = 600Psi

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

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3 years ago
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