Particle A has half the mass and eight times the kinetic energy of particle B. What is the speed ratio vA / vB?

Physics

1 answer:

Marta_Voda [28]3 years ago

8 0

Answer:

Use the equation "KE=½mv²", and use some algebra. > "Particle A has two times the mass...of particle B" mA = 2mB > "Particle A has...8 situations the kinetic power of particle B" KE_A = 8(KE_B) or: ½(mA)(vA)² = 8(½(mB)(vB)²) the rest is uncomplicated algebra: in basic terms sparkling up the above equation for "vA/vB". (hint: start up by utilising dividing the two factors by utilising "(mB)(vB)²". Then make the substitution: mA/mB = 2 (from the 1st eq0.5 ma * VA^2 = 8 * 0.5 * mb * VB^2

0.5 * 0.5*mb * VA^2 = 8 * 0.5 * mb * VB^2

0.5 * 0.5* VA^2 = 8 * 0.5 * VB^2

VA/VB = 4uation))

Explanation:

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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: