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8_murik_8 [283]
3 years ago
8

A 110 kg quarterback is running the ball downfield at 4.5 m/s in the positive direction when he is tackled head-on by a 150 kg l

inebacker moving at -3.8 m/s. Assume the collision is completely inelastic. What is the velocity of the players just after the tackle?
Physics
1 answer:
melisa1 [442]3 years ago
5 0

Answer:

v_f=-0.29\frac{m}{s}

Explanation:

The principle of conservation of momentum, states that if the sum of the forces acting on a system is null, the initial total momentum of the system before a collision equals the final total momentum of the system after the collision. The collision is completely inelastic, which means that the players remain stick to each other after the collision:

p_i=p_f\\m_1v_1+m_2v_2=(m_1+m_2)v_f\\v_f=\frac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\v_f=\frac{(110kg)4.5\frac{m}{s}+150kg(-3.8\frac{m}{s})}{(110kg+150kg)}\\v_f=-0.29\frac{m}{s}

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A thin metallic spherical shell of radius 0.357 m has a total charge of 5.03 times 10^-6 C placed on it. At the center of the sh
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Answer:

The electric field is 5.623\times10^{4}\ N/C

Explanation:

Given that,

Radius = 0.357 m

Charge Q=5.03\times10^{-6}\ C

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Distance = 0.815 m

We need to calculate the total electric field

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}

Where, q = point charge

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Put the value into the formula

E=\dfrac{9\times10^{9}\times4.15\times10^{-6}}{(0.815)^2}

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3 years ago
One relationship between organisms is that of predator-prey. Which of the following is the best description of a predator?
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4 years ago
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Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
jek_recluse [69]

Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

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         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

              θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

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b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

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whereby

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The zeros of this function are in

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           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

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