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11111nata11111 [884]
3 years ago
6

What is the circumference of the moon?

Physics
2 answers:
Fittoniya [83]3 years ago
8 0

The diameter of the moon is about 2,161 miles or (3,476 kilometers).that makes the circumference about 6,790 miles

iren [92.7K]3 years ago
7 0
The circumference of the moon is 6,786 mi
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A student pushes on a 8-kg box with a force of 35 N forward. The force of sliding friction is 10 N backward. What is the acceler
Ne4ueva [31]

Answer:

(35 N - 10 N)/8kg = 3.125 m/s^2

Explanation:

The formula for Force is:

Force = Mass*Acceleration

(Force is equal to Mass times Acceleration)

Since we're told to find the acceleration of the box. We make acceleration the subject of the equation:

Acceleration = Force/Mass

(Acceleration equal to Force divided by Mass)

We know that the force are 35 N forward and 10 N backward, and the weight of the box is 8kg.

= (35 N - 10 N)/8kg

The reason that 35 N minus 10 N is because the 10 N is pushing the box backward.

= 25 N/8kg

= 3.125 m/s^2

Hope it helps :DD

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For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
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