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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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Does the refraction of light make a swimming pool seem deeper or shallower? explain your answer.
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A common flashlight bulb is rated at 0.32 A and 4.3 V (the values of the current and voltage under operating conditions). If the
sleet_krkn [62]

Answer:

1176.01 °C

Explanation:

Using Ohm's law,

V = IR................. Equation 1

Where V = Voltage, I = current, R = Resistance when the bulb is on

make R the subject of the equation

R = V/I.................. Equation 2

R = 4.3/0.32

R = 13.4375 Ω

Using

R = R'(1+αΔθ)............................. Equation 3

Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature

make Δθ the subject of the equation

Δθ = (R-R')/αR'.................. Equation 4

Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹

Substitute into equation 4

Δθ = (13.4375-1.6)/(1.6×0.0064)

Δθ = 11.8375/0.01024

Δθ = 1156.01 °C

But,

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T₂ = T₁+Δθ

Where T₂ and T₁ = Final and initial temperature respectively.

T₂ = 20+1156.01

T₂ = 1176.01 °C

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