Figure Q1.3
The diagram does not indicate any position in time that should have been represented by numbers on the dots. Without numbers on the dots we cannot tell if the particle in the figure is moving left or right, so we can’t tell if it is speeding up or slowing down. If the particle is moving to the right it is speeding up. If it is moving to the left it is slowing down.
Part a
Zeroes before the decimal point merely locate the decimal point and are not significant. Thus, 310 has 2 significant figures.
Part b
Trailing zeros after the decimal point are significant because they indicate increased precision. Therefore, 0.00310 has 3 significant figures.
Part c
Zeroes between nonzero digits are significant. Therefore, 1.031 has 4 significant figures.
Part d
Just like in part b, trailing zeros after the decimal point are significant because they indicate increased precision.Therefore, 3.10×105 has 3 significant figures.
Part a
All nonzero digits are significant, therefore, 53.2 has 3 significant figures.
Part b
The number 0.53 can be written in scientific notation as 5.3×10-1. Therefore, 0.53 has 2 significant figures.
Part c
Trailing zeros are significant because they indicate increased precision. Therefore, 5.320 has 4 significant figures.
Part d
The leading zeros are not significant but just locate the decimal point. Therefore, 0.0532 has only 3 significant figures.
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)
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:
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*}Solution:
The known values are: y_0=1.80\:\text{m}, y=0\:\text{m}, a=-9.80\:\text{m/s}^2, v_0=4.00\:\text{m/s}.
Part A
Based from the knowns, the formula most applicable to solve for the time is \Delta y=v_0t+\frac{1}{2}at^2. If we rearrange the formula by solving for t, and substitute the given values, we have
\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2-2a\Delta y}}{a} \\
t & =\frac{-4.00\:\text{m/s}\pm \sqrt{\left(4.00\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.80\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\
t & =1.14\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part B
We have the formula
\begin{align*}
\Delta y & =\frac{v^2-v_0^2}{2a} \\
\Delta y & =\frac{\left(0\:\text{m/s}\right)^2-\left(4.00\:\text{m/s}\right)^2}{2\left(-9.80\:\text{m/s}^2\right)} \\
\Delta y & =0.816\: \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
Part C
The formula to be used is
v^2=v_0^2+2a\Delta y
Substituting the given values
\begin{align*}
v^2 & =v_0^2+2a\Delta y \\
v & =\pm \sqrt{v_0^2+2a\Delta y} \\
v & =\pm \sqrt{\left(4.00\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-1.80\:\text{m}\right)} \\
v & =\pm \sqrt{51.28\:\text{m}^2/\text{s}^2} \\
v &=\pm 7.16\:\text{m/s}
\end{align*}Since the diver must be moving in the negative direction,
v=-7.16\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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