Question

42,43 and 44 please

Answer 1

42) The sailboat travels east with velocity $v_e=30 mph$, while the current moves south with speed $v_s=30 mph$. Since the two velocities are perpendicular to each other, he resultant velocity will be given by the Pytagorean theorem:
$v= \sqrt{v_e^2+v_s^2}= \sqrt{(30)^2+(30)^2}=42 mph$
and the direction is in between the two original directions, therefore south-east. So, the correct answer is
D) 42 mph southeast

43) Since the light moves by uniform motion, we can calculate the distance corresponding to one light year by using the basic relationship between velocity, space and time. In fact, we know the velocity:
$v=300,000,000 m/s=3 \cdot 10^8 m/s$
and the time is one year, corresponding to:
$t=32,000,000 s=3.2 \cdot 10^7 s$
therefore, the distance corresponding to one light year is:
$S=vt=(3\cdot 10^8 m/s)(3.2 \cdot 10^7 s)=9.6 \cdot 10^{15}m$
Therefore, the correct answer is D.

44) For the purpose of the problem, we can assume that the light travels instantaneously from the flash to us (because the distances involved are very small), so the time between the flash and the thunder corresponds to the time it took for the sound to travel to us.
The speed of sound is
$v=1100 ft/s=750 mph=335 m/s$
And since the time between the flash and the thunder is t=3 s, the distance is
$S=vt=(335 m/s)(3 s)=1005 m=0.62 miles$
Therefore, the correct answer is A) 3/5 mile.