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2 years ago

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Rank the five terrestrial worlds in order of size from smallest to largest: Group of answer choices Mercury, Venus, Earth, Moon,

Mars. Mercury, Moon, Venus, Earth, Mars. Moon, Mercury, Venus, Earth, Mars. Moon, Mercury, Mars, Venus, Earth. Mercury, Moon, Mars, Earth, Venus.

1 answer:

dimaraw [331]2 years ago

7 0

Answer: Moon, Mercury, Mars, Venus, Earth

Explanation:

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An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi

shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = $a^{2}$

Let the height of the bin be 'h'

Therefore the total area, $A_{t} = 4ah$

The cost is:

C = 2sh

Volume of the box, V = $a^{2}h = 4^{2}\times 2 = 128$ (1)

Total cost, $C_{t} = 2a^{2} + 2ah$ (2)

From eqn (1):

$h = \frac{128}{a^{2}}$

Using the above value in eqn (1):

$C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}$

$C(a) = 2a^{2} + \frac{256}{a}$

Differentiating the above eqn w.r.t 'a':

$C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}$

For the required solution equating the above eqn to zero:

$\frac{4a^{3} - 256}{a^{2}} = 0$

$\frac{4a^{3} - 256}{a^{2}} = 0$

a = 4

Also

$h = \frac{128}{4^{2}} = 8$

The path in order to minimize the cost must be a rectangle.

8 0

3 years ago

What is the frequency of a wave that has a period of vibration of 2 seconds?

34kurt

Answer:

The answer is 0.5 Hz

Explanation:

Its pretty easy to get the answer. One hertz (Hz) is equal to one cycle or period per second. So, just divide the period by the number of seconds.

1 period/2 secs = 1/2 Hz or 0.5 Hz

7 0

2 years ago

Review the Four Social Errors and Biases in the Highlights area on page 127 in Ch. 4 of THiNK: Critical Thinking and Logic Skill

AysviL [449]

Answer: Provided in the explanation

Explanation:

I have understood that I have been influenced by the 'affinity bias' for quite a while. It caused me to feel fascination and feel better and right about individuals who had comparable intrigue and thought designs. I sort of began to feel this is the one for me based on those likenesses yet later used to be miserable when I comprehended that those similitudes are not many and insufficient consistently. I have begun to beat this inclination by rehearsing self reflection. I have begun to think about what causes me to feel pulled in to somebody and on the off chance that I introspect that it's exclusively founded on likenesses, at that point I cause myself to get that and monitoring that helps in escaping the inclination.

I make an effort not to get into the snare of paradoxes in my own announcements yet I have been forced to bear deceptions during contentions. One deception which I have encountered most is the 'Foul play' error as there have been a great deal of occurrences when individuals began to tear down me and my family when they couldn't win on a contention regarding balanced and rationale.

Basic reasoning causes us in understanding our defects and those issues of our own which keeps us from taking better choices and furthermore forestalls us tackling issues in more successful manners. Through basic reasoning, one can reflect and recognize utilization of deceptions in the contentions and that will help him in deciphering the circumstance from an alternate perspective. He will assess the circumstance better and through appropriate induction, he will have the option to get over those issues in thinking and subsequently, have the option to take better and more powerful choices for taking care of an issue.

The most fascinating idea which I have learned is that of the utilization of reflection or explicitly self reflection to comprehend ourselves better and furthermore to display basic reasoning appropriately. Being said that, I am a still somewhat confused about the manners by which we can reflect appropriately at each circumstance in a successful manner. This disarray remains in light of the fact that most deceptions and predispositions are oblivious in nature while the basic considering aptitudes reflection is cognizant and I meander whether a cognizant ability will have the option to break down and distinguish each oblivious inclinations or whether a few guards will keep a few inclinations covered up.

4 0

3 years ago

The table shows data for the planet Uranus. A 2 column table with 4 rows. The first column is labeled Quantity with entries, Esc

prohojiy [21]

Answer:

The answer is 218

Explanation:

Weight = mass * gravitational acceleration

weight is represented by F

F = 25kg (8.7)

(I'm pretty sure that you don't have to include the meters per second/per second thing)

4 0

3 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago