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3 years ago

Select the sentence from the section "An Ultracool Dwarf" that BEST explains what TRAPPIST-1 is.

SAT

1 answer:

romanna [79]3 years ago

6 0

Answer: The system’s host star, called TRAPPIST-1, is what is known as an ultracool dwarf.

Explanation:

Also called 2MASS J23062928-0502285, TRAPPIST-1 is an ultracool dwarf star that hosts three planets: TRAPPIST-1b, c, and d.

Never before planets had been found around ultracool dwarf stars, and they are considered as potentially habitable and, therefore, suitable for life outside our solar system.

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Which were british field marshal bernard montgomery contributions.

Scrat [10]

C. He defeated the Axis armies of Italy and Germany at El Alamein, Egypt.

D. He defeated the German army at the Battle of the Bulge.

7 0

3 years ago

A point charge q1 = -4. 00 nc is at the point x = 0. 60 m, y = 0. 80 m , and a second point charge q2 = +6. 00 nc is at the poin

Alekssandra [29.7K]

The net electric field is the vector sum of the components of the electric

field produced by the two charges.

The values of the magnitude and direction of the net electric field at the origin (approximate values) are;

131.6 N/C

12.6 ° above the negative x–axis

<h3>How are the net electric field magnitude and direction calculated?</h3>

The possible questions based on a similar question posted online are;

(a) The net electric field at the origin.

The electric field due to charge q₁ is given as follows;

$\vec E_{1x} = \mathbf{ \dfrac{1}{4 \cdot \pi \cdot \epsilon_0} \cdot \dfrac{q_1}{\vec{r}^2_2}}$

Which gives;

$\vec{E}_{1x} =\mathbf{ \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot cos\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) =-21.6 \, N/C$

$\vec{E}_{1y} = \mathbf{\dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot sin\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) = 28.8 \, N/C$

Which gives;

$\vec{E}_1 = \mathbf{21.6 \, N/C \cdot \hat x + 28.8 \, N/C \hat y}$

$\vec{E}_{2x} = \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(6.00 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2} = 150 \, N/C$

Therefore;

$\vec {E} = \left[ 21.6 \, N/C - 150 \, N/C \right] \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)$

$\vec {E} = \mathbf{\left( -128.4 \, N/C \right) \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)}$

The magnitude of the net electric field is therefore;

E = $\sqrt{(-128.4^2 + 28.8^2)}$ ≈ 131.6

The magnitude of the net electric field at the origin is E ≈<u> 131.6 N/C</u>

(b) The direction of the net electric field at the origin.

$The \ direction \ is \ arctan \left(\dfrac{28.8}{-128.4} \right) \approx \underline{ 12.6^{\circ}} \ above \ the \ negative \ x-axis$

Learn more about electric field strength here:

brainly.com/question/14743939

brainly.com/question/3591946

3 0

2 years ago

The accompanying histograms show the miles per gallon (mpg) for two makes of cars. Compare the distributions. Which make has a

horsena [70]

Acura cars have higher mpg while BMW has more variation in mpg.

From the histograms of mpg distribution for the two makes of car :

ACURA has most of its values at the center of the distribution with it's tail more conspicuous to the right at a maximum mpg value of 40 and a minimum of about 17.5 to the left.
BMW on the other hand has its mpg distribution more spread out towards both sides of the distribution, ranging from a minimum of about 3 mpg and to a maximum of about 37.5 mpg.
Hence, we could infer that, the histogram plot of BMW is more dispersed and hence has greater variation than ACURA.
Since ACURA has it's peak at the middle, with minimum and maximum mpg values of 17.5 and 40 respectively, which are sufficiently greater than the minimum and maximum mpg values of BMW (3.0 and 37.5 respectively). Then we can conclude that ACURA has greater mpg.

Learn more : brainly.com/question/21269371?referrer=searchResults

7 0

3 years ago

Antique Accents tracks their daily profits and has found that the distribution of profits is approximately normal with a mean of

mezya [45]

Answer;

a) 0.434

b) 0.983

c) 0.367

Explanation:

The exact question with the given parameters wasn't obtained online, but the same question, albeit with different parameters is then obtained. Hopefully, this Helps to solve the complete question with the required parameters.

Antique Accents tracks their daily profits and has found that the distribution of profis is approximately normal with a mean of $17,700.00 and a standard deviation of about $900.00. Using this information, answer the following questions For full marks your answer should be accurate to at least three decimal places. Compute the probability that tomorrow's profit will be

a) less than $16,791 or greater than $18,231

b) greater than $15,783

c) between $17,997 and $20,130

Solution

This is a normal distribution problem with

Mean = μ = $17,700

Standard deviation = σ = $900

a) less than $16,791 or greater than $18,231. P(x < 16,791) or P(X > 18,231) = P(X < 16,791) + P(x > 18,231)

We first standardize 16,791 and 18,231

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 16791

z = (x - μ)/σ = (16791 - 17700)/900 = - 1.01

For 18231

z = (x - μ)/σ = (18231 - 17700)/900 = 0.59

To determine the required probability

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

We'll use data from the normal probability table for these probabilities

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

P(z < -1.01) = 0.15625

P(z > 0.59) = 1 - (z ≤ 0.59) = 1 - 0.7224 = 0.2776

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59) = 0.15625 + 0.2776 = 0.43385 = 0.434 to 3 d.p

b) greater than $15,783. P(x > 15783)

We standardize 15783

z = (x - μ)/σ = (15783 - 17700)/900 = -2.13

To determine the required probability

P(x > 15783) = P(z > -2.13)

We'll use data from the normal probability table for this probability

P(x > 15783) = P(z > -2.13) = 1 - P(z ≤ - 2.13)

= 1 - 0.01659 = 0.98341 = 0.983 to 3 d.p.

c) between $17,997 and $20,130.

P(17,997 < x < 20,130)

We first standardize 17,997 and 20,130

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 17,997

z = (x - μ)/σ = (17,997 - 17700)/900 = 0.33

For 20,130

z = (x - μ)/σ = (20,130 - 17700)/900 = 2.70

To determine the required probability

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

We'll use data from the normal probability table for these probabilities

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

= P(z < 2.70) - P(z < 0.33)

= 0.99653 - 0.62930

= 0.36723 = 0.367 to 3 d.p.

Hope this Helps!!!

3 0

3 years ago

Read this sentence from "Cats: Entertaining at a House Near You":

inn [45]

Answer:

A. Cats give back as much love as they receive.

Explanation:

They provide companionship, often asking for attention or affection that is then returned to their human."

A. Definitely talks about this.

B. Doesn't really talk about asking a lot of attention, just attention, so not this.

C. It doesn't talk about the owners as much as it talks about the cat, so not this.

D. Dogs is not even part of the sentence so not this.

5 0

3 years ago

Read 2 more answers