Answer to Question #120361 in Physics for Grasya na ni guys

Question

1. (a) How fast must you be approaching a red traffic light (λ=675 nm) for it to appear yellow ( λ=575
nm)? Express your answer in terms of the speed of light c. (b)If you used this as a reason not to get a
ticket for running a red light, how much fine would you get for speeding? Assume that the fine is
P200.00 for each kilometer per hour that your speed exceeds the posted limit of 90 km/h.(Note: c= λf;
for (b) the solution is similar to conversion of units)
2. Calculate the magnitude of the force required to give a 0.175-kg baseball an acceleration a=1 m/s2
in
the direction of the baseball’s initial velocity when this velocity has a magnitude of (a) 10.0 m/s;
(b)0.9c; (c)0.99c. (d) Repeat parts (a), (b), and (c) if the force and acceleration are perpendicular to the
velocity.

Answer 1

Solution.

1.$\lambda_0=675nm=675\sdot10^{-9}m;$

$\lambda=575nm=575\sdot10^{-9}m;$

$c=3\sdot10^{8}m/s;$

a)$f=\sqrt{\dfrac{c+u}{c-u}}f_0;$

$f^2=\dfrac{c+u}{c-u}f_0^2;$

$f^2(c-u)=f_0^2(c+u);$

$f^2c-f^2u=f_0^2c+f_0^2u;$

$f^2c-f_0^2c=f^2u+f_0^2u;$

$u=c\dfrac{f^2-f_0^2}{f^2+f_0^2};$

$f=\dfrac{c}{\lambda}, f_0=\dfrac{c}{\lambda_0};$

$f=\dfrac{3\sdot10^8m/s}{575\sdot10^{-9}m}=5.2\sdot10^{14}Hz;$

$f_0=\dfrac{3\sdot10^8m/s}{675\sdot10^{-9}m}=4.4\sdot10^{14}Hz;$

$u=3\sdot10^8\dfrac{(5.2\sdot10^{14}Hz)^2-(4.4\sdot10^{14}Hz)^2}{(5.2\sdot10^{14}Hz)^2+{(4.4\sdot10^{14}Hz)^2}}=0.165c(m/s);$

b)$u=0.165c=0.165\sdot3\sdot10^3m/s=4.96\sdot10^7m/s=1.79\sdot10^8km/h;$

$P=(1.79\sdot10^8-90)\sdot200.00=356\sdot10^8$ - the fine;

2.$m=0.175kg;$

$a=1m/s^2;$

$F=ma;$

$F=0.175kg\sdot1m/s^2=0.175N;$

Answer: 1. a)$u=0.165c(m/s);$

$P=356\sdot10^8$ - the fine;

2.$F=0.175N;$