How long did it take the flag to rotate once in a full circle

Can someone please explain to me why the net force of the ball changes in the Y direction and not the X Direction? I made little

stiks02 [169]

Whenever an object is in projectile motion, that is, it has 2-dimensional motion in the x and y axis, the resultant force on the object is in the y-direction.
This is because once the object has been projected, or the ball has been kicked in this case, there is no longer a force being applied on it in the x-direction. The air resistance is also neglected so the ball's final velocity in the x-direction is equal to its initial velocity in the x-direction.
However, the force of gravity cannot be neglected and causes the ball to come downwards. Therefore, after the ball has been projected, the net force on the ball is downwards, due to gravity.

8 0

3 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: