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2 years ago

What does the word self discipline mean to you

2 answers:

6 0

I think being self disciplined means you know right from wrong. You don’t need other people always telling you what is right and what you do wrong because you should know it yourself. I think this means you’re also well behaved and you know when to stop doing something if it pushes someone to the edge. Two habits I would associate with being self disciplined would be to make up your bed every morning or if someone says to stop doing something, you stop.

sorry if this is wrong :(

posledela2 years ago

3 0

I think being self disciplined is when you are a responsible and mature person and maybe raised yourself I don’t know what else to write you can expand on this

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Genrish500 [490]

The blue line with triangles are a cold front. The red line with half circles are a warm front. The black H is high pressure. The black L is low pressure. The black dot is overcast skies. Finally the white circle with a line down and 4 lines off of that line is 40-knot wind

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8 0

2 years ago

What is the speed of a bobsled whose distance-time graph indicates that it traveled 100 m in 25 s?

larisa86 [58]

As per the question, the distance travelled by bobsled [s] = 100 m

The time taken by the bobsled to travel that distance [t] = 25 s

We are asked to calculate the speed of the bobsled.

The speed of the bobsled is calculated as -

$speed\ =\ \frac{distance}{time}$

$v\ =\ \frac{s}{t}$

$=\frac{100\ m}{25\ s}$

$=\ 4\ m/s$

Hence, the correct answer to the question is A. 4 m/s.

8 0

3 years ago

How do I solve the following problem?

lukranit [14]

Use the impulse-momentum theorem.

$Ft = mv$

Substitute your known values:

$(120kg)(20m/s) = F(1.5)

F= 1600N$

Hope this helps!

7 0

3 years ago

Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera

torisob [31]

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

No intersection: There's nothing that blocks the camera's view of the top of the building.
Two intersections: The planet blocks the camera's view of the top of the building.
One intersection: The point at which the top of the building appears or disappears.

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle $\angle \mathrm{B\hat{C}D}$ which corresponds to this minor arc.

This angle comes can be split into two parts:

$\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}$ .

Also,

$\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}$ .

The radius of this circle is:

$\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}$ .

The lengths of segment DC, AC, BC can all be found:

$\rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m$ ;
$\rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m$ ;
$\rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m$ .

In the two right triangles $\triangle\mathrm{DAC}$ and $\triangle \rm BAC$ , the value of $\angle \mathrm{B\hat{C}A}$ and $\angle \mathrm{A\hat{C}D}$ can be found using the inverse cosine function:

$\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}$

$\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}$

$\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}$ .

The length of the minor arc will be:

$\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km$ .

5 0

2 years ago

The angular separation of two stars is 0.1 arcseconds and you photograph them with a telescope that has an angular resolution of

gizmo_the_mogwai [7]

Answer:

Will see them as only one star

Explanation:

Solution:

- The angular resolution of a telescope means the minimum quantity that can be visualized. Since their angular separation ( 0.1 arcseconds ) is smaller than the telescope's angular resolution (1 arcseconds ), your photograph will seem to show only one star rather than two.

7 0

2 years ago