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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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Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

F_{att}=G\frac{m_{1}m_{2}}{r^{2}}

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N

Force of attraction between 435 kg mass and 38 kg mass is

F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N

Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

8 0
3 years ago
What is the kinetic energy of a 12 kg ball moving at a speed of 15 m/s?
vivado [14]

Answer:

1350 N

Explanation:

KE = ½mv²

KE = ½(12)(15²)

KE = 1350 N

3 0
1 year ago
An object is dropped from rest from a height of 4.4 107 m above the surface of the Earth. If there is no air resistance, what is
kondaur [170]

Answer:

Explanation:

An object is drop from a height, then it is in direction of gravity

g is +ve

When an object drop from a height, the initial velocity is 0,

U=0

Given that h=4.4×10^7m

V=?

g=9.81m/s^2

Then using equation of motion

V^2=U^2+2gh

V^2=0+2×9.81×4.4×10^7

V^2=86.33×10^7m/s

Take square root of both side

V=29381.63m/s

Now to km/s, divide by 1000

Since 1km=1000m

V=29.38km/s

6 0
3 years ago
How does temperature affect the spring constant of a spring? Detailed answer pls. :)
trasher [3.6K]

Answer by, Kkmmll44 :P

Converting the measured force into a resultant displacement requires an accurate spring constant. This spring constant may change as the temperature rises due to microstructural effects. As the temperature increases, the spring constant decreases. A linear fit was made to the data to determine the spring constant

Equations!

F = kx. F is the force applied to the spring in newtons (N) k is the spring constant measured in newtons per meter (N/m) x is the distance the spring is stretched from its equilibrium position in meters (m) The Hooke's Law Apparatus.

Effects!

The heat treatment process alters the microstructure and strength of material, which will directly influence fatigue life [2]. Spring manufacturing requires materials

Brainliest please :)

Hope this helps!

3 0
2 years ago
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Nellie Newton drops her shoe out of a helicopter traveling parallel to the ground at 80m/s. The shoe falls to the ground 6 secon
Nostrana [21]

Answer:

480 m

Explanation:

If the shoe is falling for 6 seconds and the helicopter is falling at 80 m/s you would mutiply 6*80=480

hope this helped !

7 0
3 years ago
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