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Mariulka [41]
3 years ago
9

An explosive projectile is launched straight upward to a maximum height h. At its peak, it explodes, scattering particles in all

directions. Assume each particle has the same initial speed v after the explosion. What is the smallest angle that the final velocity of a particle will make with the horizontal upon hitting the ground
Physics
1 answer:
Varvara68 [4.7K]3 years ago
3 0

Answer:

θ = tan⁻¹ (\frac{19.6 \ h}{v})

Explanation:

This problem must be solved using projectile launch ratios. Let's analyze the situation, the projectile explodes at the highest point, therefore we fear the height (i = h), the speed at this point is the same, but the direction changes, we are asked to find the smallest angle of the speed in the point of arrival with respect to the x-axis.

The speed at the arrival point (y = 0)

           v² = vₓ² + v_y²

Let's see how this angle changes, for two extreme values:

* The particle that falls from the point of explosion, in this case the speed is vertical

         v = v_y

the angle with the horizontal is 90º

* The particle leaves horizontally from the point of the explosion, the initial velocity is horizontal

         vₓ = v

the final velocity for y = 0

         v_f = vₓ² + v_y²

therefore the angle has a value greater than zero and less than 90º

As they ask for the smallest angle, we can see that we must solve the last case

the output velocity is horizontal vₓ = v

Let's find the velocity when it hits the ground y = 0, with y₀ = h

            v_{y}^2 = v_{oy}^2 - 2 g (y-y₀)

            v_{y}^2 = - 2g (0- y₀)

let's calculate

           v_{y}^2 = 2 9.8 h

         

we use trigonometry to find the angle

        tan θ = \frac{v_y}{v_x}

        θ = tan⁻¹ (\frac{v_y}{v_x})

let's calculate

         θ = tan⁻¹ (\frac{19.6 \ h}{v})

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To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

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Here,

M = Mass

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The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

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d = Distance of center of mass from pivot

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Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

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T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

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An empty office chair is at rest on a floor. Consider the following forces:. 1. A downward force due to gravity;. 2. An upward forc
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 4. 1 and 2 only.

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The mass of a spacecraft is about 435 kg . An engine designed to increase the speed of the spacecraft while in outer space provi
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Δ v =  125 m/s

Explanation:

given,

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thrust = 0.09 N

time = 1 week

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change in speed of craft = ?

Assuming no external force is exerted on the space craft

now,

T= m_s a

a=\dfrac{T}{m_s}

a =\dfrac{0.09}{435}

a = 2.068 x 10⁻⁴ m/s²

using equation of motion

Δ v = a t

Δ v = 2.068 x 10⁻⁴ x 7 x 24 x 60 x 60

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3 0
3 years ago
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Answer:

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Now, with our value of maximum electromagnetic wave gotten, we divide it by speed of light to get our final answer

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I think it’s D-decreases the amount of work.

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