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krek1111 [17]
2 years ago
11

Your friend (68 kg) is wearing frictionless roller skates and standing still. You throw at her a 3.6 kg pumpkin with a velocity

of 9.5 m/s and she catches it. What will be her resultant velocity ?
Physics
1 answer:
Fudgin [204]2 years ago
6 0

Answer:

Approximately 0.48\; \rm m\cdot s^{-1}.

Explanation:

Momentum would be conserved since there's no friction on this friend, and all other forces on her are balanced. Therefore:

\begin{aligned}& \text{Resultant momentum of (friend and pumpkin)} \\ =\; & \text{Initial momentum of (friend)} \\ & + \text{Initial momentum of (pumpkin)}\end{aligned}.

Momentum p the product of mass m and velocity v. That is: p = m \, v.

The initial momentum of this friend is 0\; \rm kg \cdot m \cdot s^{-1} since she was initially not moving (an initial velocity of 0\; \rm m\cdot s^{-1}.)

The initial momentum of the pumpkin would be:

\begin{aligned}p &= m \, v \\ &= 3.6 \; \rm kg \times 9.5\; \rm m\cdot s^{-1} \\ &= 34.2\; \rm kg \cdot m \cdot s^{-1}\end{aligned}.

Therefore:

\begin{aligned}& \text{Resultant momentum of (friend and pumpkin)} \\ =\; & \text{Initial momentum of (friend)} \\ & + \text{Initial momentum of (pumpkin)} \\ =\; &0\; {\rm kg \cdot m \cdot s^{-1}} + 34.2\; {\rm kg \cdot m \cdot s^{-1}} \\ =\; & 34.2\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}.

Rearrange the equation p = m \, v to find an expression for velocity v given momentum and mass:

\displaystyle v = \frac{p}{m}.

Note that the "final momentum of friend and pumpkin" in the previous equation refers to the resultant velocity of the friend with the pumpkin in her hand. Thus, it would necessary to use the combined mass of the friend and the pumpkin (68\; {\rm kg} + 3.6 \; {\rm kg}) when calculating the resultant velocity:

\begin{aligned}& \text{Resultant velocity of (friend and pumpkin)} \\ =\; & \frac{\text{Resultant momentum of (friend and pumpkin)}}{\text{Mass of (friend and pumpkin)}} \\ =\; & \frac{34.2\; {\rm kg \cdot m \cdot s^{-1}}}{68\; {\rm kg} + 3.6\; {\rm kg}} \\ \approx \; & 0.48\; \rm m \cdot s^{-1}\end{aligned}.

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Answer:

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Explanation:

For this exercise we will use the law of universal gravitation

          F = - \frac{m_{1} m_{2} }{r^2}

We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m

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          F₁ = - \frac{m' M }{r^2}

rocket force - Moon

          F₂ = - \frac{m' m }{(d-r)^2}

in the problem ask for what point the force has the relation

          2 F₁ = F₂

let's substitute

          2 2 \frac{M}{r^2} = \frac{m}{(d-r)^2}

          (d-r) ² = \frac{m}{2M} r²

           d² - 2rd + r² = \frac{m}{2M} r²

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Let's solve this quadratic equation to find the distance r, let's call

           a = 1 - \frac{m}{2M}

           a = 1 - \frac{7.36 10^{22} }{2 \  5398 10^{24}} = 1 - 6.15 10⁻³

           a = 0.99385

         

            a r² - 2d r + d² = 0

           r =  \frac  {2d \frac{+}{-}   \sqrt{4d^2 - 4 a d^2}} {2a}

           r = [2d ± 2d \sqrt{1-a}] / 2a

           r = \frac{d}{a}   (1 ± √ (1.65 10⁻³)) =  \frac{d}{a} (1 ± 0.04)

           r₁ = \frac{d}{a} 1.04

           r₂ = \frac{d}{a} 0.96

let's calculate

           r₁ = \frac{3.84 10^8}{0.99385} 1.04

           r₁ = 401.8 10⁸ m

          r₂ = \frac{3.84 10^8}{0.99385} 0.96

          r₂ = 3.71 10⁸ m

therefore the correct result is r = 3.71 10⁸ m

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3 years ago
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Answer:

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We have to determine the energy of the incident electron.

Mass of electron,m=9.1\times 10^{-31} kg

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Using the formula

Speed of electron,v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0

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Answer:

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Width of one fringe = \frac{\lambda\times D}{d}

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= 2 times 4.86

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b) width of each fringe except central fringe  is same , no matter what the order is.Only brightness changes .

So width of either of the two first order bright fringe will be same and it will be  

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Answer:

Inverse Square Law Newton proposed the Inverse Square Law. The effect of gravity (and also on forces such as sunlight) works like this. If say we have a half-mass Earth, it would produce a gravity of not half but a quarter (the square of 2).

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