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2 years ago

12

(See picture) may I have help!!?

1 answer:

algol132 years ago

6 0

Picture is blurry…. try re uploading it

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Why doesn’t Florida get as much earthquakes as California does

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2 years ago

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3 track athletes run the 400 meter dash, Jackie 51 seconds and 1500 joules

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2 years ago

A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point

seropon [69]

since centripetal acceleration is always towards the center of the circle

so at the given position where speed and acceleration is given the center coordinate will be towards the center of circle

also we know that

$a_c = \frac{v^2}{r}$

$11.6 = \frac{2.5^2}{R}$

$R = 0.54 m$

so the coordinates of the center will be

$(x, y) = (3.20 , (3.5 + 0.54))$

$(x, y) = (3.20, 4.04)$

so the coordinate is (3.20 m, 4.04 m)

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2 years ago

In your own words, explain what conservation of energy means. Also, give an example of the conservation of energy using somethin

Genrish500 [490]

Energy can not be created or destroyed but can change from one form to another.

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2 years ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

3 0

3 years ago