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Feliz [49]
3 years ago
13

an archer stands 40.0m from the target. if the arrow is shot horizontally with a velocity of 90.0 m/s, how far above the bull's

eye must she aim to compensate for gravity pulling her arrow downward? ...?
Physics
1 answer:
pshichka [43]3 years ago
3 0
Ok, first we need to have in mind that to findthe distance we need to find also the time then we use the kinematic equations in the x-direction 

<span>delta x = Vox*t + 1/2*a*t^2 </span>
<span>since there is no horizontal acceleration, a = 0 m/s/s </span>
<span>delta x = Vox*t </span>
<span>40.0 m = 90 m/s * t </span>
<span>t = 4/9 s </span>

<span>Now using this time you can calculate the vertical displacement due to gravity. </span>
<span>delta y = Voy*t + 1/2*g*t^2 </span>
<span>since there is no initial vertical velocity, Voy = 0m/s </span>
<span>delta y = 1/2*g*t^2 </span>
<span>delta y = 1/2*(-9.81 m/s/s)*(4/9 s)^2 </span>
<span>solve for delta y and you will have the answer</span>
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Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when
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<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

v₁ is the final speed of car X.

m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s

So, car X will move with a velocity of -12 m/s.

3 0
3 years ago
Please help me fill out the chart.??
LekaFEV [45]

Answer:unbalanced: have direction,Change an objects motion, causes object to accelerate

Balanced:Do not change an objects motion, Net forces equal sum of all forces on object, and Does not equal 0 N

Explanation: Thats all I know Hoped I helped sum

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