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Alex787 [66]
3 years ago
15

The woman was standing on a cliff 30 meters above where she landed. There was a safety fence 4 meters from the cliff edge that w

ould have prevented her from taking a long distance to run and leap off the cliff. She landed 12 meters away from the base of the cliff. Her launch speed was calculated to be 4.86 m/s, meaning that to travel those 12 meters, she had to be traveling at 10.87 miles per hour (or moving faster than a runner at a 5.5 minute mile pace) as she left the cliff. Put all this information together. Why did forensic scientists determine that the woman’s body was thrown instead of concluding that she jumped off the cliff?
Physics
1 answer:
pav-90 [236]3 years ago
4 0

Answer:

Forensic scientists determined that the woman’s body was thrown off the cliff because after considering the velocity and angle of the projectile needed in relationship to the cliff she could not have reached those speeds and cleared the safety fence 4m away from the cliff’s edge. She would have been going faster than a 5.5 minute mile pace and she was not an athletic person.

** Just make sure to change the format and/or wording. Hope I was helpful :-)

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Before the internet was available people often search for jobs?
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A revolutionary war cannon, with a mass of 2090 kg, fires a 16.7 kg ball horizontally. The cannonball has a speed of 113 m/s aft
OLEGan [10]

Answer:

0.90291 m/s

0.45055 m/s

Explanation:

m_1 = Mass of canon = 2090 kg

m_2 = Mass of ball = 16.7 kg

v_1 = Velocity of canon

v_2 = Velocity of ball = 113 m/s

In this system the momentum is conserved

m_1v_1=m_2v_2\\\Rightarrow v_1=\dfrac{16.7\times 113}{2090}\\\Rightarrow v_1=0.90291\ m/s

The velocity of the cannon is 0.90291 m/s

Applying energy conservation

\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2=\dfrac{1}{2}m_2v^2\\\Rightarrow m_1v_1^2+m_2v_2^2=m_2v^2\\\Rightarrow v_2=\sqrt{\dfrac{m_1v_1^2+m_2v_2^2}{m_2}}\\\Rightarrow v_2=\sqrt{\dfrac{2090\times 0.90291^2+16.7\times 113^2}{16.7}}\\\Rightarrow v_2=113.45055\ m/s

The ball would travel 113.45055-113 = 0.45055 m/s faster

6 0
4 years ago
A car is driving northwest at v mph across a sloping plain whose height, in feet above sea level, at a point n miles north and e
coldgirl [10]
Refer to the diagram shown.

Given:
h(n,e) = 1500 + 75n + 50e

Define 
\hat{r} = unit \, vector \, along \, \vec{v} \\ \hat{i} = unit \, vector \, east \\ \hat{j} = unit \, vector \, north \\ \nabla \equiv \hat{i}  \frac{\partial}{\partial e} + \hat{j}  \frac{\partial}{\partial n}

\hat{r} =  \frac{1}{ \sqrt{2} } (-\hat{i}+\hat{j} )

Then the rate of change of h with respect to the vector v is
\nabla h . \hat{r} = \frac{1}{\sqrt{2}}(50\hat{i} + 75\hat{j}).(-\hat{i}+\hat{j}) = \frac{1}{\sqrt{2}} (-50+75) =17.68

Answer: 17.7 ft per mile

3 0
3 years ago
4-What determines the length of a planet's day? *
shusha [124]

Answer:

Axis rotation to determine the length of a day

Explanation:

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7 0
4 years ago
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