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Eva8 [605]
3 years ago
9

Skateboard A Force: 33 N Time: 0.12 s Skateboard B Force: 33 N Time: 0.57 s Skateboard C Force: 33 N Time: 0.78 s Which skateboa

rd has the greatest impulse?
Physics
2 answers:
nata0808 [166]3 years ago
7 0
The impulse is given by the magnitude of the force times the time interval:
I= F \Delta t
We can see that the force F of the 4 skateboard is always the same (33 N), this means that the skateboard with greatest impulse is the one with the largest time interval, so skateboard C).
abruzzese [7]3 years ago
7 0
It is C. just the letter C. do not type skateboard c.
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Explanation:

Given that,

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(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2meE}}

\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

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(b) According to Bragg's law,

n\lambda=2d\ sin\theta

n = 1

For nickel, d=0.092\times 10^{-9}\ m

\theta=sin^{-1}(\dfrac{\lambda}{2d})

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\theta=0.010^{\circ}

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

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3 years ago
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v_f=0.825m/s

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We must use conservation of linear momentum before and after the collision, p_i=p_f

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After the collision we have:

p_f=(m_1+m_2)v_f

since they clong together, acting as one body.

This means we have:

m_1v_1+m_2v_2=(m_1+m_2)v_f

Or:

v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}

Which for our values is:

v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s

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Explanation:

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