How do you determine the label for each vertex when you perform the triangle triangle in problem 5

Two balls of equal size are dropped from the same height from the roof of a building. One ball has twice the mass of the other.

frutty [35]

Answer:

The kinetic energy of the more massive ball is greater by a factor of 2.

Explanation:

By conservation of energy, we know that the initial energy = final energy. At first, the balls are dropped from a height with no initial velocity so their initial energy is all potential energy. When they reach the bottom, all their energy is kinetic energy. So all of their energy is changed from potential to kinetic energy. This means that the ball with greater potential energy will have a greater kinetic energy.

Potential energy = mgh. Since g = gravity is a constant and h = height is the same, the only difference is mass. Since mass is directly proportional to potential energy, the greater the mass, the greater the potential energy, so the more massive ball has a greater initial potential energy and will have a greater kinetic energy at the bottom.

Additionally, let B1 = lighter ball with mass m and let B2 = heavier ball with mass m2. Since we know that intial potential energy = final kinetic energy. We can rewrite it as potential energy = kinetic energy = mass * gravity constant * height. For B1, it is mgh and for B2 it is 2mgh, so B2's kinetic energy is twice that of B1.

3 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: