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andriy [413]
1 year ago
14

The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviatio

n of 0.01 inches. What percentage of bolts will have a diameter greater than 0.32​ inches?
Physics
1 answer:
enot [183]1 year ago
8 0

Answer:

2.25 %

Explanation:

65-95-99.7 is a rule to remember the precentages that lies around the mean.

at the range of mean (\mu) plus or minus one standard deviation (\sigma), P([\mu-\sigma \leq X \leq \mu+\sigma])\approx 68.3%

at the range of mean plus or minus two standard deviation, P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%

at the range of mean plus or minus three standard deviation, P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%

So, note that  they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) (P(X \leq \mu+2\sigma))  

So we know that the 95.5% is between \mu - 2\sigma = 0.3 -2*0.1 = 0.28 and \mu + 2\sigma = 0.3 +2*0.1 = 0.32, hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than  0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.

So P(X \leq \mu+2\sigma) \approx 2.25%

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Explanation:

Since the temperature is constant, the gas would expand isothermally.

For isothermal expansion,

P₁V₁=P₂V₂

Where, P₁ and P₂ are the initial and final pressure and V₁ and V₂ are initial and final volume.

It is given that:

V₁ = 0.0250 m³

P₁ = 1.5 × 10⁶ Pa

P₂ = 0.950 × 10⁶ Pa

V₂ = ?

⇒ 1.5 × 10⁶ Pa × 0.0250 m³ = 0.950 × 10⁶ Pa × V₂

⇒V₂ = 0.039 m³

Hence, the final volume V₂ of the container is  0.039 m³.

4 0
1 year ago
How does end A of the rod react when the charged ball approaches it after a great many previous contacts with end A? Assume that
sleet_krkn [62]

Answer: IT IS STRONGLY REPELLED

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The rod will be strongly repelled because the forces on the rod is greater and has the same Polarity as the charged ball.

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Para proteger un computador de sobrecargas eléctricas, Juan coloca un filamento delgado de cobre llamado fusible en su circuito,
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Answer:

Los fusibles están diseñados de tal forma que estos se "rompen" o se funden, cuando la demanda eléctrica supera un dado valor (cuando demasiada electricidad pasa a través de el).

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Entonces, al romperse el filamento, en caso de una sobrecarga eléctrica, el flujo de electricidad se corta, y de esta forma se protege al computador de posibles sobrecargas.

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1 year ago
Levi and Clara are trying to move a very heavy box. Levi is pushing the box with a force of 30 N, and Clara is pulling the box w
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There is a ner force of 15 N allowing Levi and Clara to mobe the box.
5 0
2 years ago
Read 2 more answers
 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer
Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}

we know that y_{f}=0 and that v_{y0}=0 as well, so the formula is simplified to:

0=y_{0}+\frac{1}{2}at^{2}

we can now solve this for t, so we get:

-y_{0}=\frac{1}{2}at^{2}

-2y_{0}=at^{2}

\frac{-2y_{0}}{a}=t^{2}

t=\sqrt{\frac{-2y_{0}}{a}}

we know that y_{0}=1.20m and that a=g=-9.8m/s^{2}

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

V_{x}=\frac{x}{t}

so we can solve this for x, so we get:

x=V_{x}t

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

a=\frac{v_{f}-v_{0}}{t}

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

a=\frac{v_{f}}{t}

so we can solve this for the final velocity:

V_{yf}=at

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

V_{yf}=(-9.8m/s^{2})(0.495s)

which yields:

V_{yf}=-4.851m/s

so now we know the components of the final velocity, which are:

V_{xf}=1.5m/s and V_{yf]=-4.851m/s

so now we can find the speed by determining the magnitude of the vector, like this:

|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}

so we get:

|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

\theta = tan^{-1} (\frac{V_{y}}{V_{x}})

so we get:

\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})

which yields:

\theta = -72.82^{o}

4 0
2 years ago
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