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3 years ago

The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur.

Engineering

1 answer:

OlgaM077 [116]3 years ago

3 0

Answer:

a) the velocity of the implant immediately after impact is 20 m/s

b) the average resistance of the implant is 40000 N

Explanation:

a) The impulse momentum is:

mv1 + ∑Imp(1---->2) = mv2

According the exercise:

v1=0

∑Imp(1---->2) = F(t2-t1)

m=0.2 kg

Replacing:

$0+F(t_{2} -t_{1} )=0.2v_{2}$

if F=2 kN and t2-t1=2x10^-3 s. Replacing

$0+2x10^{-3} (2x10^{-3} )=0.2v_{2} \\v_{2} =\frac{4}{0.2} =20m/s$

b) Work and energy in the system is:

T2 - U(2----->3) = T3

where T2 and T3 are the kinetic energy and U(2----->3) is the work.

$T_{2} =\frac{1}{2} mv_{2}^{2} \\T_{3} =0\\U_{2---3} =-F_{res} x$

Replacing:

$\frac{1}{2} *0.2*20^{2} -F_{res} *0.001=0\\F_{res} =40000N$

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A body weighs 50 N and hangs from a spring with spring constant of 50 N/m. A dashpot is attached to the body. If the body is rai

lbvjy [14]

Answer:

a) 3.607 m

b) 1.5963 m

Explanation:

See that attached pictures for explanation.

3 0

2 years ago

Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is

Leona [35]

Answer:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

$dQ = dU - dW$

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) $\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW$

As per work definition:

$dW = F*dr$

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

$dW = PA*dx$

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

$dW = - P*dV$

So the third integral in equation (1) is:

$\int\limits^1_2 {- P} \, dV$

Considering the gas as ideal, the pressure can be calculated as $P = \frac{n*R*T}{V}$ , so:

$\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV$

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

$\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

Replacion this and solving equation (1) between state 1 and 2:

$\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}$

The internal energy depends only on the temperature of the gas, so there is no internal energy change $U_{2} - U_{1} = 0$ , so the heat exchanged to the system equals the work done by the system:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

4 0

3 years ago

A_____ transducer is a device that can convert an electronic controller output signal into a standard pneumatic output. A. pneum

makkiz [27]

Answer:

The correct answer is

option C. current to pneumatic (V/P)

Explanation:

A current to pneumatic controller is basically used to receive an electronic signal from a controller and converts it further into a standard pneumatic output signal which is further used to operate a positioner or control valve. These devices are reliable, robust and accurate.

Though Voltage and current to pressure transducers are collectively called as electro pneumatic tranducers and the only electronic feature to control output pressure in them is the coil.

6 0

3 years ago

3<br> Current is measured in units called

bixtya [17]

Answer:

current is measured in Ampere (A)

6 0

2 years ago

Read 2 more answers

Cho P1= XdaN, P2=3.X daN, P= 2.X (daN). a=1m, b=2m,MCN hình tròn d= (100+X)mm1/ Tính nội lực tại các mặt cắt cách ngàm 0,5m; 1m;

DaniilM [7]

Answer: