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zepelin
1 month ago
9

Rutherford used positively charged particles to investigate the structure of the atom. The results surprised him, and he develop

ed the atomic model shown below. What surprising result is explained using this model? A few positive particles bounced back because they bounced off of the net formed by the ovals. A few positive particles bounced back because they were pushed away from the positive center. Most positive particles passed through because of the empty space between the outer edge and the center. Most positive particles passed through because they were attracted by the small particles moving around the center.
Mathematics
2 answers:
Zina [11.9K]1 month ago
6 0

Answer: The accurate statement is Most positive particles passed through because of the empty space found between the edges and the center.

Explanation:

Ernest Rutherford conducted what is known as the Gold Foil Experiment.

In this experiment, he used a gold foil and bombarded it with alpha particles, which have a positive charge of +2. Initially, he expected that many particles would reflect back, but he was astonished to find that an overwhelming number passed through, with some deviating and a few bouncing back.

This led him to deduce that within an atom, a small positive charge is concentrated at the center, while an atom is predominantly made up of empty space.

Thus, the surprising conclusion is that most positive particles passed through due to the empty space between the outer edges and the nucleus.

Zina [11.9K]1 month ago
3 0

The answer is (B): A few positive particles were reflected because they were repelled by the positive core.

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Svet_ta [12230]

Answer:

0.40 feet

Step-by-step explanation:

For the first scenario, a 50-foot building casts a shadow of 1 foot. Let the angle of elevation of the sun from the shadow be denoted as θ.

Then:

Tan θ = \frac{opposite}{adjacent}

Tan θ = \frac{50}{1}

Tan θ = 50

⇒ θ = Tan^{-1} 50

      = 88.8542

      = 88.85^{o}

The elevation angle is roughly 88.85^{o}.

For a 20-foot pole,

Tan θ = \frac{opposite}{adjacent}

Tan 88.85^{o} = \frac{20}{x}

x = \frac{20}{Tan 88.85^{o} }

 = 0.4015

 = 0.40 feet

The length of the pole's shadow is 0.40 feet.

4 0
10 days ago
A bank advertises an APR of 5.5% on personal loans. How much more is the APY when the rate is compounded monthly as compared to
tester [11849]
0.027%. A bank promotes an APR of 5.5% for personal loans. To address this problem, we will utilize the Annual Percentage Yield formula. In this formula, r signifies the interest rate in decimal form, and n represents the number of compounding periods per year. First, we convert the interest rate into decimal format. Next, we will calculate APY while compounding monthly using n = 12 and r = 0.055 within the APY formula. We proceed to do the same for quarterly compounding by substituting n = 4 and r = 0.055 into the APY formula. To determine the difference, we subtract the quarterly APY from the monthly APY. Therefore, the APY for monthly compounding is 0.027% higher than for quarterly compounding.
3 0
3 days ago
Mike runs for the president of the student government and is interested to know whether the proportion of the student body in fa
zzz [11803]

Answer:

We conclude that less than or equal to 50% of the student body supports him significantly.

Step-by-step explanation:

Mike, while campaigning for the student government presidency, is keen to determine if more than 50% of the student body supports him.

A random sample of 100 students was surveyed, with 55 expressing support for Mike.

Let p = the proportion of students backing Mike.

Thus, Null Hypothesis, H_0: p \leq 50% {indicating that the proportion of supporters among the student body is significantly less than or equal to 50%}

Alternate Hypothesis, H_A: p > 50% {indicating that the proportion of supporters among the student body is significantly more than 50%}

The test statistics to be utilized here One-sample z proportion statistics;

T.S. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } ~ N(0,1)

where, \hat p = the sampled proportion in favor of Mike = \frac{55}{100} = 0.55

n = number of sampled students = 100

Therefore, test statistics = \frac{0.55-0.50}{\sqrt{\frac{0.55(1-0.55)}{100} } }

= 1.01

The z test statistic is 1.01.

At a significance level of 0.05, the z table provides a critical value of 1.645 for a right-tailed test.

Given that our test statistic falls below the critical value of z (1.01 < 1.645), we lack sufficient evidence to reject the null hypothesis as it remains outside the rejection region, leading to failure to reject our null hypothesis.

As a result, we conclude that less than or equal to 50% of the student body is in favor of him or the proportion supporting Mike is not significantly greater than 50%.

4 0
1 month ago
A professor believes that students at her large university who exercise daily perform better in statistics classes. Since all st
Leona [12066]

Response: Yes, it is

Detailed explanation:

Since the average scores of both sets of students vary, evaluating the mean score in relation to their class will clarify which class supports or contradicts the professor's research

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12 days ago
The simplest form of 1.666….. is?
Leona [12066]
El valor más simple sería 5/3 en forma de fracción.
3 0
1 month ago
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