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2 years ago

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Use Gauss's Law to find the charge contained in the solid hemisphere x2 + y2 + z2 ≤ a2, z ≥ 0, if the electric field is E(x, y,

z) = x i + y j + 6z k.

1 answer:

Anit [1.1K]2 years ago

6 0

Answer:<em><u> $\frac{16}{3}$ π $a^{3}$ . </u></em>

Given:

$x^{2}+y^{2}+z^{2} \leq a^{2}, z \geq 0$

Using Gauss's Law = ∫∫s E ·dS

= ∫∫∫ div E dV,

⇒ Divergence (Gauss') Theorem

= ∫∫∫ (1+1+6) dV

= 8×(volume of the hemisphere, radius "a")

= 8× ( $\frac{1}{2}$ )(4/3)π $a^{3}$

<em><u>= $\frac{16}{3}$ π $a^{3}$ . </u></em>

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Two triangles are similar with a scale factor of 3. The preimage triangle has a base of 7 inches and a height of 10 inches. What

vaieri [72.5K]

Answer:

<u>The area of the larger triangle is 315 inches²</u>

Step-by-step explanation:

1. Let's review all the information provided for answering the questions properly:

Pre-image triangle has a base of 7 inches and a height of 10 inches

Scale used : Factor of 3

2. What is the area of the larger triangle??

For calculating the area of the larger triangle, we use the scale this way:

Base = 7 inches * 3

21 inches

Height = 10 inches * 3

30 inches

Area of the larger triangle = 1/2 (Base * Height)

Area of the larger triangle = 1/2 (21 * 30)

<u>Area of the larger triangle = 630/2 = 315 inches²</u>

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2 years ago

If a boy and a half can eat a hot dog and a half, in a minute and a half, how many hot dogs can six boys eat in six minutes?

Yanka [14]

SIx boys can eat seven (7) hotdogs in six (6) minutes.

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2 years ago

Mustafa, Heloise, and Gia have written more than a combined total of 2222 22 articles for the school newspaper. Heloise has writ

Papessa [141]

Answer: $m>8$

Mustafa could have written more than 8 articles for the newspaper.

Step-by-step explanation:

Given: Heloise has written 1/4 as many articles as Mustafa has. Gia has written 3/2 as many articles as Mustafa has.

Let m be the number of articles Mustafa writes.

Then articles written by Heloise= $\frac{1}{4}m$

And articles written by Gia= $\frac{3}{2}m$

The total number of articles = $m+\frac{1}{4}m+\frac{3}{2}m=\frac{4m+m+6m}{4}=\frac{11m}{4}$

Mustafa, Heloise, and Gia have written more than a combined total of 22 articles for the school newspaper.

$\frac{11m}{4}>22\\\Rightarrow\ m>\frac{22\times4}{11}\\\Rightarrow\ m>8$

Hence, Mustafa could have written more than 8 articles for the newspaper.

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2 years ago

Read 2 more answers

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

klio [65]

Answer:

a) The mean is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.

This means that $\mu = 10, \sigma = 2$

Suppose 64 visitors independently view the site.

This means that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and the variance of the mean time of the visitors.

Using the Central Limit Theorem, mean of 10 and variance of (0.25)^2 = 0.0625.

b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, so between 9.75 and 10.25 seconds, which is the p-value of Z when X = 10.25 subtracted by the p-value of Z when X = 9.75.

X = 10.25

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{10.25 - 10}{0.25}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

X = 9.75

$Z = \frac{X - \mu}{s}$

$Z = \frac{9.75 - 10}{0.25}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

0.8413 - 0.1587 = 0.6826.

0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c. The value exceeded by the mean time of the visitors with probability 0.01.

The 100 - 1 = 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.

$Z = \frac{X - \mu}{s}$

$2.327 = \frac{X - 10}{0.25}$

$X - 10 = 2.327*0.25$

$X = 10.58$

So 10.58 minutes.

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2 years ago

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Ad libitum [116K]

Answer:

yay! sabrina is making an open box

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2 years ago