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3 years ago

15

A runner jogs 1 km south to a school, then jogs 1km west to a restaurant. What is the position of the runner relative to the sch

ool

1 answer:

Finger [1]3 years ago

7 0

Answer:

12

Explanation:

yeah

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When does the fall equinox occur in the northern hemisphere

aivan3 [116]

Here is your answer

On 23rd September

HOPE IT IS USEFUL

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2 years ago

Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

nydimaria [60]

Answer: $F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

$F_1=\frac{kq(-q)}{(L\sqrt{2})^2}$

where $L\sqrt{2}=$ Distance between the two charges

$F_1=-\frac{kq^2}{2L^2}$

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

$F_2=\frac{kq(-q)}{(L)^2}$

The magnitude of force by both the charge is same but at an angle of $90^{\circ}$

thus combination of two forces at 2 and 3 will be

$F'=\sqrt{2}\frac{kq^2}{2L^2}$

Now it will add with force due to 1 charge

Thus net force will be

$F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

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2 years ago

It is known as a basic unit of all living things.

andrezito [222]

hello, the answer is..

cell

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2 years ago

Rearrange the formula F=ma, and solve for the variable (a)

jeyben [28]

Answer:

a = F/m

Explanation:

So we have to isolate a, in order to do this we need to move m to the other side, and we do that by diving both sides by m, resulting in a = F/m

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2 years ago

slader the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an a

Ostrovityanka [42]

Answer:

0.08 ft/min

Explanation:

To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.

So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth <em>d</em> and the other (lets call it <em>l</em>) is given by:

$l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft$

since the difference between the upper and lower base is the increase in the base and we are only at halft the height.

Now we can calculate the longitudinal section <em>A</em> at that point:

$A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}$

And the raising speed <em>v </em>of the water is given by:

$v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}$

where <em>q</em> is the water flow (1 cubic foot per minute).

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3 years ago