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2 years ago

A fast-moving car has 0.5 MJ of kinetic energy.

Physics

1 answer:

DerKrebs [107]2 years ago

6 0

Answer:

-0.5 MW

Explanation:

For this we can use the principle of work and energy, which states that:

T₁ + ∑U₁₋₂ = T₂

Where T is kinetic energy (in situations one and two) and U is the sum of all work done (between situations one and two).

So, since the car is first moving with a kinetic energy of 0.5 MJ:

T₁ = 0.5 MJ = 0.5 · 10⁶ J

On top of this, we know that the car comes to a halt, using this as situation 2:

T₂ = 0 J

Filling in the equation:

⇒ 0.5 · 10⁶ + U = 0

⇒ U = -0.5· 10⁶ [W] (unit of work is watts)

= -0.5 MW

Thinking about the sign convention, we see that the brake force will point to the back of the car (backward), and the distance the car will travel before stopping will be forward, oposite directions and thus negative work.

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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: