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3 years ago

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A football player kicks the ball at a 45 degree angle. Without an effect from the wind, the ball would travel 60.0m horizontally

. A) what is the initial speed of the ball? B) when the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally from the initial kick point? (In m)

1 answer:

Ronch [10]3 years ago

5 0

B when the ball is at its maxium height

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

ELEN [110]

With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$

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A child of mass 25 kg is skating fast, +10 m/s, and tries to get revenge by colliding with the 60 kg adult who is sitting still.

tankabanditka [31]

Answer: -4.4 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1}V_{o}+m_{2}U_{o}$ (2)

$p_{f}=m_{1}V_{f}+m_{2}U_{f}$ (3)

$m_{1}=25 kg$ is the mass of the child

$V_{o}=10 m/s$ is the initial velocity of the child

$m_{2}=60 kg$ is the mass of the adult

$U_{o}=0 m/s$ is the initial velocity of the adult (it is sitting still)

$V_{f}$ is the final velocity of the child

$U_{f}=6 m/s$ is the final velocity of the adult

Substituting (2) and (3) in (1):

$m_{1}V_{o}+m_{2}U_{o}=m_{1}V_{f}+m_{2}U_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1}V_{o}-m_{2}U_{f}}{m_{1}}$ (5)

$V_{f}=\frac{(25 kg)(10 m/s)-(60 kg)(6 m/s)}{25 kg}$ (6)

Finally:

$V_{f}=-4.4 m/s$ This means the velocity of the child is in the opposite direction

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