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
You can browse on the itemized questions with solutions of the College Physics by Openstax below. Also, you can buy the whole Complete Solution Manual here.
Chapter 14: Heat and Heat Transfer Methods
Chapter 15: Thermodynamics
Chapter 16: Oscillatory Motion and Waves
Chapter 17: Physics of Hearing
Chapter 18:
Electric Charge and Electric Field
Chapter 19:
Electric Potential and Electric Field
Chapter 20:
Electric Current, Resistance, and Ohm’s Law
Chapter 21: Circuits and DC Instruments
Chapter 22:
Magnetism
Chapter 23:
Electromagnetic Induction, AC Circuits, and Electrical Technologies
Chapter 24:
Electromagnetic Waves
Chapter 25: Geometric Optics
Chapter 26: Vision and Optical Instrument
Chapter 27: Wave Optics
Chapter 28: Special Relativity
Chapter 29: Introduction to Quantum Physics
Chapter 30: Atomic Physics
Chapter 31: Radioactivity and Nuclear Physics
Chapter 32: Medical Applications of Nuclear Physics
Chapter 33:
Particle Physics
Chapter 34: Frontiers of Physics
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)
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