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Strike441 [17]
2 years ago
7

The small 5-oz slider A moves without appreciable friction in the hollow tube, which rotates in a horizontal plane with a consta

nt angular speed of ω=7 rad/sec. The slider is launched with an initial speed of dr=60 ft/sec relative to the tube at the inertial coordinates x=6 in. and y=0. Determine the agnitude P of the horizontal force exerted on the slider by the tube just before the slider exits the tube.

Engineering
1 answer:
amid [387]2 years ago
7 0

Answer:

Magnitude of P=8.624 lb

Explanation:

The complete solution to the problem is given in the attachments.

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The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 3.7 × 104 148 1.0 × 105 13
prohojiy [21]

Answer:

i) S–N plot is attached

ii) fatigue strength = 100 MPa

iii) fatigue life = 5.62 x 10^(5) cycles

Explanation:

i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

Log(4 × 10^(6)) = 6.6

From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa

iii) The question says we should find the fatigue life for 120 MPa.

Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

Thus,the fatigue life will be the inverse log of 5.75.

Thus, fatigue life = 10^(5.75)

Fatigue life = 5.62 x 10^(5)

8 0
2 years ago
A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a
Fed [463]

Answer:

<em>minimum required diameter of the steel linkage is 3.57 mm</em>

<em></em>

Explanation:

original length of linkage l = 10 m

force to be transmitted  f = 2 kN = 2000 N

extension e = 5 mm= 0.005 m

maximum stress σ = 200 N/mm^2 = 2*10^{8}  N/m^{2}

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/(2*10^{8}) = 10^{-5} m^2

recall that area = \pi d^{2} /4

10^{-5} = \frac{3.142*d^{2} }{4} = 0.7855d^{2}

d^{2} = \frac{10^{-5} }{0.7855} = 1.273*10^{-5}

d = \sqrt{1.273*10^{-5}  } = 3.57*10^{-3} m = 3.57 mm

<em>maximum diameter of  the steel linkage d = 3.57 mm</em>

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Who want to play "1v1 lol unblocked games 76"
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Answer:I want to play what game?

Explanation:

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Your task is to fill in the missing parts of the C code to get a program equivalent to the generated assembly code. Recall that
Rudik [331]

Answer:

See Explaination

Explanation:

//Function

long loop (long x, long n)

{

//Declare a variable named result and initialize it to zero

long result = 0;

//Declare a variable named mask

long mask;

//For loop

for(mask = 1; mask != 0; mask = mask << (n & 0xFF))

{

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result | = (x&mask);

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6 0
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At his review last year, Lucas was promised a 20 percent raise if he met his production goals. Raises were included in today’s p
Murrr4er [49]

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