SOLUTION:
Draw the free-body diagram of the car
Consider the vertical direction
Consider the motion in the horizontal direction
Solve for the acceleration of the car.
Solve for the coefficient of kinetic friction
Solution:
The known values are a=-9.80\:\text{m/s}^2; v_o=15.0\:\text{m/s}; y=7.00\:\text{m}
The applicable formula is.
y=v_ot+\frac{1}{2}at^2Using this formula, we can solve it in terms of time, t.
t=\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a}Substituting the known values, we have
\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a} \\
t & =\frac{-15.0\:\text{m/s}\pm \sqrt{\left(15.0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(7.00\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\
t&=\frac{-15.0\:\text{m/s}\pm 9.37\:\text{m/s}}{-9.80\:\text{m/s}^2}
\end{align*}We have two values for time, t. These two values represent the times when the ball passes the tree branch.
t_1=\frac{-15.0\:m/s+9.37\:m/s}{-9.80\:m/s^2}=0.57\:sec \\
t_2=\frac{-15.0\:m/s-9.37\:m/s}{-9.80\:m/s^2}=2.49\:secTherefore, the total time between passing the branch is the difference between 2.49 seconds and 0.57 seconds.
t_2-t_1=2.49 \ \text{s} - 0.57 \ \text{s}=1.92 \ \text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Displacement
Time, Velocity, and Speed
Acceleration
Motion Equations for Constant Acceleration in One Dimension
Falling Objects
Graphical Analysis of One-Dimensional Motion
Physical Quantities and Units
Accuracy, Precision, and Significant Figures
Approximation
The formula for the total distance traveled is
\Delta s=\Delta \theta \times r
Therefore, the total distance traveled is
\begin{align*}
\Delta s & =\left(200000\:\text{rotations}\:\times \frac{2\pi \:\text{radian}}{1\:\text{rotation}}\right)\left(\frac{1.15\:\text{m}}{2}\right) \\
\Delta s & =722566.3103\:\text{m} \\
\Delta s & =722.6\:\text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Solution:
The formula for friction is
f=\mu _{k\:}NWhen we solve for the normal force, N, in terms of the other variables, we have
N=\frac{f}{\mu _k}The coefficient of kinetic friction is 0.04. Therefore, the normal force is
\begin{align*}
N & =\frac{f}{\mu _k} \\
N & =\frac{0.200\:\text{newton}}{0.04} \\
N & =5.00\:\text{newton} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Solution:
So, we are given mass, m = 63.0 \ \text{kg} , and acceleration, a = 4.20 \ \text{m/s}^2.
The net force has a formula
\text{F}=\text{m}aSubstituting the given values, we have
\begin{align*}
F & = \left( 63.0 \ \text{kg} \right)\left( 4.20 \ \text{m/s}^2 \right) \\
F & = 265 \ \text{kg}\cdot \text{m/s}^2 \\
F & = 265 \ \text{N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Solution:
Part A
The total distance traveled is
\begin{align*}
\text{d} & =\left(3\times 120\ \text{m}\right)+\left(1\times 120\:\text{m}\right) \\
\text{d} & =480\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}Part B
The magnitude of the displacement is
\begin{align*}
\text{s }& =\sqrt{\left( s_x \right)^{2\:}+\left( s_y \right)^2} \\
\text{s }& = \sqrt{\left(1\times 120\:\text{m}\right)^2+\left(3\times 120\:\text{m}\right)^2} \\
\text{s }& = 379\ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}The direction is
\begin{align*}
\theta & = \tan^{-1}\left(\frac{s_x}{s_y}\right) \\
\theta & = \tan^{-1}\left(\frac{1\times 120\:\text{m}}{3\times 120 \ \text{m}}\right) \\
\theta & =71.6^{\circ} ,\:\text{E of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Solution:
The known values are: y_0=2.20\:\text{m}; y=1.80\:\text{m}; v_0=11.0\:\text{m/s}; and a=-9.80\:\text{m/s}^2
We are going to use the formula
\Delta y=v_0t+\frac{1}{2}at^2Substituting the given values:
\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
1.80\:\text{m}-2.20\:\text{m} & =\left(11.0\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\
-0.40\:\text{m} & =\left(11.0\:\text{m/s}\right)t-\left(4.90\:\text{m/s}^2\right)t^2 \\
4.90t^2-11t-0.40 & =0
\end{align*}Using the quadratic formula solve for t, we have
\begin{align*}
t & =\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\left(4.90\right)\left(-0.40\right)}}{2\left(4.90\right)} \\
\end{align*} t=2.28\:\text{sec}\:\:\:\:\:\text{or}\:\:\:\:\:\:t=-0.04 \ \text{sec}We can discard the negative time, so
t=2.28\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)Solution:
Part A
Refer to the figure below.
The known values are: t=2.35\:\text{s}; y=0\:\text{m}; v_0=+8.00\:\text{m/s}; and a=-9.8\:\text{m/s}^2
Based on the given values, the formula that we shall use is
y=y_0+v_0t+\frac{1}{2}at^2Substituting the values, we have
\begin{align*}
y & =y_0+v_0t+\frac{1}{2}at^2 \\
0\: & =y_0+\left(8.00\:\text{m/s}\right)\left(2.35\:\text{s}\right)+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(2.35\:\text{s}\right)^2 \\
y_0 & =8.26\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}Therefore, the cliff is 8.26 meters high.
Refer to the figure below
The knowns now are: y=0\:\text{m}; y_0=8.26\:\text{m}; v_0=-8.00\:\text{m/s}; and a=-9.80\:\text{m/s}^2
Based on the given values, we can use the formula
y=y_0+v_0t+\frac{1}{2}at^2Substituting the values, we have
\begin{align*}
y & =y_0+v_0t+\frac{1}{2}at^2 \\
0\:\text{m} & =8.26\:\text{m}+\left(-8.00\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\
4.9 t^2+8t-8.26 & =0 \\
\end{align*}
Using the quadratic formula to solve for the value of t, we have
\begin{align*}
t &=\frac{-8\pm \sqrt{\left(8\right)^2-4\left(4.9\right)\left(-8.26\right)}}{2\left(4.9\right)} \\
t &=0.717\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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