Do all of them please and explain

Physics

2 answers:

kondaur [170]3 years ago

5 0

At 305 and 309 it was negative because it is going down

at 300 and 301 and 305 it was positive because its going up

at 302/303/304/306/309 she wasn't accelerating at all because it is staying the same

k0ka [10]3 years ago

5 0

Okay, so for 11: She is negatively accelerating between 3:05 and 3:06 and between 3:08 and 3:09

When you have a speed vs distance graph, the slope of the graph represents the acceleration, since there is a negative slope there we can say she was negatively accelerating

12. So now all we have to do it look for a line going up or a positive slope:

So: 3:00 to 3:02 , 3:07 to 3:08

13. So, when we are looking for no acceleration we are looking for a straight line, or a line with no slope at all which is between the intervals of:

3:02 to 3:05 , 3:06 to 3:07 and 3:09 to 3:10

Hope this helped, let me know if any of it doesnt make sense.

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a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

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$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$