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Breakup of a Kaon at Rest
Motion in One Dimension • 1.1 Position, Velocity, and Speed
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Breakup of a Kaon at Rest
The velocity of a particle moving along the $x$ axis varies in time according to the expression $v_x = (40-5t^2)$, where $t$ is in seconds
Find the average acceleration in the time interval between $t=0.0$ to $t=2.0$ seconds.
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keyboard_arrow_downWe find the velocities at $t_i=t_A=0$ and $t_f=t_B=2.0 \ s$ by substituting these values of $t$ into the expression for the velocity:
$$v_{xA}=(40-5t_A^2) \ m/s = [40 - 5(0)^2] \ m/s = +40 \ m/s$$$$v_{xB}=(40-5t_B^2) \ m/s = [40 - 5(2.0)^2] \ m/s = +20 \ m/s$$
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Physics II: Electricity & Magnetism
10 units • 780 problems
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keyboard_arrow_downThe charge enclosed by the Gaussian surface is given by
$$q_{in}= \int \rho_0 dV = \int _0^r \rho_0 4\pi r'^2dr'=\rho (\frac{4}{3}\pi r^3)$$
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Question 3
Solve the following integral for $x$
$$\int xsin(x^2)dx$$
Choose answer
A
$-\frac{1}{2} cos(x^2)$
B
$-cos(x^2)$
C
$2xcos(x^2)$
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