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3 years ago

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A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the sol

id cylindrical shaft BC. Design specifications require that the displacement of D should not exceed 15 mm from the time the punch first touches the plastic sheet to the time it actually penetrates it. Determine the required diameter of shaft BC if the shaft is made of a steel with G.

1 answer:

olasank [31]3 years ago

4 0

Answer:

hello your question lacks some data and required diagram

G = 77 GPa, т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P ) = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅ , where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅ of CD on the shaft BC

Torque = 660 * 0.3 cos∅

= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

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In a liquid with a density of 1400 kg/m3 , longitudinal waves with a frequency of 370 Hz are found to have a wavelength of 8.40

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Answer:

Bulk modulus = 1.35 × $10^{10}$ Pa

Explanation:

given data

density = 1400 kg/m³

frequency = 370 Hz

wavelength = 8.40 m

solution

we get here bulk modulus of the liquid that is

we know Bulk Modulus = $v^2*\rho$ ...............

here $\rho$ is density i.e 1400 kg/m³

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v = 370 × 8.40 = 3108 m/s

so here bulk modulus will be as

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4 years ago

An object of mass 25kg is falling from the height h=10 m. calculate

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Answer:

a=2500J,b=1000K,c=1000J,d=14.142m/s

Explanation:

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V²=0 + 2×10×10=200m/s

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b).mgh=25×10×4=1000J

c). V²=U²+2gh

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4 years ago

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A multimeter in an RL circuit records an rms current of 0.600 A and a 50.0-Hz rms generator voltage of 110 V. A wattmeter shows

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Answer:

(a) The impedance in the circuit is $Z=183.33\Omega$ .

(b)The resistance is $R=38.89\Omega$ .

(c) The inuctance is 0.57 H.

Explanation:

(a)

The expression for the impedance is as follows:

$Z=\frac{V_rms}{I_rms}$

Here, $V_rms$ is the rms voltage and $I_rms$ is the rms current.

Put $V_rms=110 V$ and $I_rms=0.600 A$ .

$Z=\frac{110}{0.600}$

$Z=183.33\Omega$

Therefore, the impedance in the circuit is $Z=183.33\Omega$ .

(b)

The expression for the average power is as follows;

$P_{a}=I_{rms}^{2}R$

Here, $P_{a}$ is the average power and R is the resistance.

Calculate the resistance by rearranging the above expression.

$R=\frac{P_{a}}{I_{rms}^{2}}$

Put $P_{a}=14W$ and

$R=\frac{14}{{0.600}^{2}}$

$R=38.89\Omega$

Therefore, the resistance is $R=38.89\Omega$ .

(c)

The expression for the impedance is as follows;

$Z^{2}=R^{2}+X_{L}^{2}$

Here, $X_{L}$ is the inductive reactance.

Put $Z=183.33\Omega$ and $R=38.89\Omega$ .

$(183.33)^{2}=(38.89)^{2}+X_{L}^{2}$

$X_{L}=179.16\Omega$

The expression for the inductive reactance in terms of frequency is as follows;

$X_{L}=2\pi fL$

Here, L is the inductance.

Calculate the inductance by rearranging the above expression.

$L=\frac{X_{L}}{2\pi f}$

Put $X_{L}=179.16\Omega$ and f=50Hz.

$L=\frac{179.16}{2\pi (50)}$

L=0.57 H

Therefore, the inuctance is 0.57 H.

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