Recent content by nmnna

I think my assumptions eariler were wrong, ##\vec{v}## does not have the same magnitude as ##\vec{v}_{\text{E}}##. The angular velocity of the plane must be equal to the angular velocity of the Earth, then the Sun will be the center of the circle relative to which the Earth with the plane is...

##v = (R_{\text{E}} + h) \cdot \frac{\pi}{12}##

I think so

We know that ##v = \omega r## where ##r = R_{\text{E}} + h##. To compensate for the motion, the plane must fly along the equator at the same speed as the Earth but in the opposite direction, i.e. from east to west, so $$\vec{v} = -\vec{ v}_{\text{E}}$$ $$v_{\text{E}} = \omega_{\text{E}}...

As I understand from your answer my solution should look something like this, right? $$\frac{\Delta{x_2}}{\Delta{x_1}} = \frac{(2/3\times12) + 3}{(9\times1) - 1} = \frac{11}{8}$$ [edit] This is wrong

I thought about this situation too, but by solving this i got ##\frac{12}{2/3\times12+3} = \frac{12}{11} \approx 1.09##, which is not the expected answer...

The graph: 1) $$v_1 = \frac{\Delta{x}}{t} = \frac{5 - 3}{3} = \frac{2}{3}$$ $$2 = \frac{\Delta{x}}{t} = \frac{4 - 0}{5 - 1} = 1$$ $$\frac{v_2}{v_1} = \frac{1}{2/3} = \frac{3}{2} = 1.5$$ 2)Points of intersection of the lines with the x-axis: ##I## (0; 3) and ##II## (0; -1), thus $$\frac{2}{3}t...

Thank you for your help.

I changed my diagram. Now I have the right triangle ##\triangle PQC##, where ##CP = 4##cm (since ##PQ## is a perpendicular bisector), ##\angle QCP = 59^{\circ}40'##, so I can find ##PQ## using the relation $$\tan\angle QCP = \frac{PQ}{CP}$$ I got ##\approx 6.818## which is not the answer given...

Thank you

Here is my attempt to draw a diagram for this problem: I'm confused about the "the perpendicular bisector of ##BC## cuts ##BA##, ##CA## produced at ##P, \ Q##" part of the problem. How does perpendicular bisector of ##BC## cut the side ##CA##?

It should be ##ACB##

The Figure My Attempt at Solution ##\tan{ACB} = \frac{AB}{BC}, \ \tan41.45^\circ = \frac{AB}{10} \Rightarrow AB = 10\tan45.41^\circ \approx 8.83##cm Similarly ##\tan{CBD} = \frac{CD}{BC}, \ \tan32.73^\circ = \frac{CD}{10} \Rightarrow CD = 10\tan32.73^\circ \approx 6.43##cm After this I...

Okey, sorry for the trouble