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Dennis_Churaev [7]
3 years ago
15

The specification limits for a product are 8 cm and 10 cm. A process that produces the product has a mean of 9.5 cm and a standa

rd deviation of 0.2 cm. What is the process capability, Cpk
Physics
1 answer:
My name is Ann [436]3 years ago
4 0

Answer:

The value of Cpk is 0.83.

Explanation:

Given that,

Upper specification limits = 10 cm

lower specification limits = 8 cm

Mean = 9.5

Standard deviation = 0.2 cm

We need to calculate the process capability

Using formula of Cpk

Cpk=min(\dfrac{USL-mean}{3\times SD}, \dfrac{mean-LSL}{3\times SD})

Put the value into the formula

Cpk=min(\dfrac{10-9.5}{3\times0.2}, \dfrac{9.5-8}{3\times0.2})

Cpk=min(0.83,2.5)

Cpk=0.83

Hence, The value of Cpk is 0.83.

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Three cars collide, a 1500 kg sports car, a 1750 kg family car, and a 1200 kg compact car. Which experiences the greatest change
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Answer:

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Explanation:

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4 0
3 years ago
Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padd
Free_Kalibri [48]

Answer:

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

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v_f = 0.87 m/s

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If each pull-up requires 300 J and Ben does a pull-up in 2 seconds, what is his power? 150 watts 300 watts 600 watts 750 watts
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Answer:

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6 0
2 years ago
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