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vitfil [10]
3 years ago
9

A golf ball strikes a hard, smooth floor at an angle of 28.2 ° and, as the drawing shows, rebounds at the same angle. The mass o

f the ball is 0.0260 kg, and its speed is 42.8 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)
Physics
1 answer:
WITCHER [35]3 years ago
7 0

Answer:

    I = 1.06886  N s

Explanation:

The expression for momentum is

          I = F t = Δp

therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor

Let's find the components of the initial velocity

          sin 28.2 = v_y / v

          cos 28.2=  vₓ / v

          v_y = v sin 282

          vₓ = v cos 28.2

          v_y = 42.8 sin 28.2 = 20.225 m / s

          vₓ = 42.8 cos 28.2 = 37.72 m / s

since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making

θ = -28.2

         v_y = -20.55 m / s

         v_x = 37.72 m / s

X axis

         Iₓ = Δpₓ = p_{fx} - p_{ox}

since the ball moves in the x-axis without changing the velocity, the change in moment must be zero

         Δpₓ = m v_{fx} - m v₀ₓ = 0

          v_{fx} = v₀ₓ

therefore

          Iₓ = 0

Y axis  

        I_y = Δp_y = p_{fy} -p_{oy}

when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards

         v_{fy} = - v_{oy}

         Δp_y = 2 m v_{oy}

         Δp_y = 2 0.0260 (20.55)

         \Delta p_{y} = 1.0686 N s

the total impulse is

          I = Iₓ i ^ + I_y j ^

          I = 1.06886  j^ N s

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Airida [17]

The equation of the wave travelling along the +x-axis is y = 0.02 sin (880π/330 x – 880 πt)

<u>Explanation:</u>

Given data

Amplitude 0.02 m , Frequency= 440 Hz ,Speed = 330 m/s

The equation  format is written as,

y =   A   sin (       k        x  –   ω     t)

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we get A value from the given data

A = 0.02 m

By the formula,

y =   A   sin (       k        x  –   ω     t)

Substitute the values we get the equation,

y = 0.02 sin (880π/330 x – 880 πt)

The equation of the wave travelling along the +x-axis is y = 0.02 sin (880π/330 x – 880 πt)

8 0
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he potential energy between two atoms in a particular molecule has the form U(s) = 2.6/x^8 - 4.3/x^4 where the units of x are le
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Answer:

x = 1.04866

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Force can be defined from power energy by the expressions

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in this case we are the expression of the potential energy

          U = \frac{2.6}{x^{8} }  - \frac{4.3}{ x^{4} }

let's find the derivative

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we substitute

          F = + \frac{20.8}{ x^{9} }  - \frac{17.2 }{ x^{5} }

at the equilibrium point the force is zero, so

           \frac{20.8}{ x^{9} } = \frac{17.2}{ x^{5} }

           20.8 / 17.2 = x⁴

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<h2><u>Question</u>:-</h2>

A 10cm³ sample of copper has a density of 897g/cm³. What is the mass of the sample ?

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<h3>Given:-</h3>

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Volume of sample of copper is 10cm³.

<h3>To Find:-</h3>

The mass of the sample.

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We know,

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