All categories

3 years ago

PLEASE HELP I NEED TO TURN IT IN IN AN HOUR ITLL GIVE YOU POINTS PLS PLEASE

Physics

1 answer:

miv72 [106K]3 years ago

5 0

Answer:

17.95025
172.3995

Explanation:

$\mathrm{The\:solution\:for\:Long\:Division\:of}\:\frac{2.995}{0.16685}\:\\\mathrm{is}\:17.95025$

$\mathrm{Multiply\:without\:the\:decimal\:points,\:then\:put\:the\:decimal\:point\:in\:the\:answer}\\\\910\times\:18945=17239950\\\\910\mathrm{\:has\:}0\mathrm{\:decimal\:places}\\0.18945\mathrm{\:has\:}5\mathrm{\:decimal\:places}\\\\\mathrm{Therefore,\:the\:answer\:has\:}5\mathrm{\:decimal\:places}\\\\=172.39950\\\\Refine\\=172.3995$

You might be interested in

Question :

irakobra [83]

A rubber ball and a stone of the same size are examples which will have more inertia and is therefore denoted as option A.

<h3>What is Inertia?</h3>

This is referred to as the property exhibited by a body in which it has the tendency to remain at rest or in uniform motion.This property is dependent on the mass of the substance as we can deduce that the greater the mass, the greater the inertia and vice versa.

The size of a rubber ball and stone will have different masses in which that of the stone will be greater. This is as a result of the difference in the nature of the substances which are used to make both items mentioned above.

This is therefore the reason why a rubber ball and a stone of the same size as having more inertia(mass) where chosen as the most appropriate choice in this scenario.

Read more about Inertia here brainly.com/question/1140505

#SPJ1

5 0

2 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:

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