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Korvikt [17]
3 years ago
15

URGENT!!! DUE AT 11:59, PLEASE DON’T POST A LINK FOR THIS ANSWER AND BE STRAIGHT FORWARD! Bronco the skydiver, whose mass is 100

kg experiences 200 N of air resistance. What is the acceleration of his fall? Show Work
Physics
1 answer:
kirill [66]3 years ago
4 0

a = 7.8 m/s^2

Explanation:

Let Fnet = net force = ma

m = mass of the skydiver

a = acceleration caused by Fnet

W = weight = mg

f(air) = frictional force due to air resistance

Fnet = W - f(air)

= (100 kg)(9.8 m/s^2) - (200 N)

= 780 N

Therefore, the acceleration of the skydiver due to Fnet is

a = Fnet/m

= (780 N)/(100 kg)

= 7.8 m/s^2

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Answer: Some challenges that I could see would be the walls, possibly a moat, tar, and the towers.

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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
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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:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} (5)

Where:

a=-4.9 m/s^{2

b=2.267 m/s

c=0.44 m

Substituting the known values:

t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)} (6)

Solving (6) we find the positive result is:

t=0.609 s (7)

Substituting (7) in (1):

x=(3.10 m/s)cos(47\°)(0.609 s)  (8)

We finally find the horizontal distance traveled by the venom:

x=1.289 m  

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