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ivolga24 [154]
3 years ago
7

Mickey, a daredevil mouse of mass m , m, is attempting to become the world's first "mouse cannonball." He is loaded into a sprin

g-powered gun pointing up at some angle, and is shot into the air. The gun's spring has force constant k k and is initially compressed a distance X X from its relaxed position. If Mickey has a constant horizontal speed V V while he is flying through the air, how high above his initial location in the gun does Mickey soar? Find an expression for Mickey's maximum height h m hm in terms of m , m, V , V, X , X, k , k, and g .
Physics
2 answers:
Sati [7]3 years ago
8 0

Answer:

  h = v₀² / 2g ,      h = k/4g     x²

Explanation:

In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches

Starting point. Lower compressed spring

              Em₀ = K = ½ m v²

Final point. Highest on the path

             Em_{f} = U = mg h

             

As or no friction the energy is conserved  

              Em₀ =  Em_{f}

              ½ m v₀²² = m g h

             h = v₀² / 2g

We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse

                  ½ k x² = 2 g h

                  h = k/4g     x²

quester [9]3 years ago
6 0

Answer:

 h = v₀² / 2g ,      h = k/4g     x²

Explanation:

edg

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When nasa launched the kepler spacecraft in 2009 what was its primary mission
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The Kepler mission is specifically designed to survey a portion of our region of the Milky Way galaxy<span> to discover dozens of Earth-size planets in or near the </span>habitable zone<span> and determine how many of the billions of stars in our galaxy have such planets</span>
3 0
3 years ago
The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
sattari [20]

Answer:

 k = 11,564 N / m,   w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

 let's apply the equilibrium condition at this point

                 

Axis y

          W_{y} - Fr = 0

          Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

             sin 46 = W_{y} / W

             W_{y} = W sin 46

     

 we substitute

           mg sin 46 = k y

           k = mg / y sin 46

If the length of the bar is L

          sin 46 = y / L

           y = L sin46

 

we substitute

           k = mg / L sin 46 sin 46

           k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

           k = 1.18 9.8 / 1

           k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

          Em₀ = U = mgy

end point. Point at 30º

         Em_{f} = K -Ke = ½ I w² - ½ k y²

          em₀ = Em_{f}

          mgy = ½ I w² - ½ k y²

          w = √ (mgy + ½ ky²) 2 / I

the height by 30º

           sin 30 = y / L

           y = L sin 30

           y = 0.5 m

the moment of inertia of a bar that rotates at one end is

          I = ⅓ mL 2

          I = ½ 1.18 12

          I = 0.3933 kg m²

let's calculate

          w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

          w = 6.06 rad / s

7 0
3 years ago
How do i do a Wavelength and frequency problem
KengaRu [80]

Answer:

wavelength = v/f or wavelength equals to velocity over frequency

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7 0
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The motor of a washing machine rotates with a period of 28 ms. What is the angular speed, in units of rad/s?
pentagon [3]

Answer:

2π/[28 x (10^-3)]

Explanation:

Angular speed : ω=2π/T

T = 28ms = 28 x (10^-3) s

Angular speed = 2π/[28 x (10^-3)]

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puteri [66]
A an objects weight hope this helps! :D 
3 0
3 years ago
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