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alexandr402 [8]
3 years ago
10

A baseball is thrown upward from a height of 2 meters with an intial velocity of 10 meters per second . determine its maximum he

ight, using a(t)=-9.8 meters per second as acceleration due to gravity.
Physics
1 answer:
Vladimir79 [104]3 years ago
6 0
<span>The equation to be used for this would be as follows:

h(x) = -9.8x^2 + 10x + 2

We need to find for the axis of symmetry for the time it takes to reach the maximum height and is expressed as:

</span><span>x = -b/(2a)
x = -10/2(-9.8)
x = 0.51 seconds is the time it takes to reach the max. height

</span>h(x) = -9.8x^2 + 10x + 2
h(x) = -9.8x0.51^2 + 10x0.51 + 2
h(x) = 4.55 m
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Answer:

a)t=1.4\times 10^{-5}\ s

b)S= 46.4 cm

Explanation:

Given that

Velocity = 16 Km/s

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d= 0.225 m

a)

lets time taken by electron is t

d = V x t

0.225 = 16,000 t

t=1.4\times 10^{-5}\ s

b)

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m=9.1\times 10^{-31}\ kg

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So now by putting the values in equation 1

a=\dfrac{E q}{m}

a=\dfrac{1.6\times 10^{-19}\times 0.027}{9.1\times 10^{-31}}\ m/s^2

a=4.74\times 10^{9}\ m/s^2

S= ut+\dfrac{1}{2}at^2

Here initial velocity u= 0 m/s

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Answer:

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Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

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Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

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        Em₀ = Em_f

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          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

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this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

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we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

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we look for the acerleracion

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       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

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     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

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a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

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