To determine the distance of the light that has traveled given the time it takes to travel that distance, we need a relation that would relate time with distance. In any case, it would be the speed of the motion or specifically the speed of light that is travelling which is given as 3x10^8 meters per second. So, we simply multiply the time to the speed. Before doing so, we need to remember that the units should be homogeneous. We do as follows:
distance = 3x10^8 m/s ( 8.3 min ) ( 60 s / 1 min ) = 1.494x10^11 m
Since we are asked for the distance to be in kilometers, we convert
distance = 1.494x10^11 m ( 1 km / 1000 m) = 149400000 km
Answer:
a) 2.41 km
b) 38.8°
Questions c and d are illegible.
Explanation:
We can express the displacements as vectors with origin on the point he started (0, 0).
When he traveled south he moved to (-3, 0).
When he moved east he moved to (-3, x)
The magnitude of the total displacement is found with Pythagoras theorem:
d^2 = dx^2 + dy^2
Rearranging:
dy^2 = d^2 - dx^2
![dy = \sqrt{d^2 - dx^2}](https://tex.z-dn.net/?f=dy%20%3D%20%5Csqrt%7Bd%5E2%20-%20dx%5E2%7D)
![dy = \sqrt{3.85^2 - 3^2} = 2.41 km](https://tex.z-dn.net/?f=dy%20%3D%20%5Csqrt%7B3.85%5E2%20-%203%5E2%7D%20%20%3D%202.41%20km)
The angle of the displacement vector is:
cos(a) = dx/d
a = arccos(dx/d)
a = arccos(3/3.85) = 38.8°
Answer:
16.8 lb is the force on the brake pad of one wheel.
Explanation:
Force applied on the piston = ![F_1=5.6 lb](https://tex.z-dn.net/?f=F_1%3D5.6%20lb)
Area of the piston = ![A_1=0.6 inches^2](https://tex.z-dn.net/?f=A_1%3D0.6%20inches%5E2)
Force applied on the brakes = ![F_2](https://tex.z-dn.net/?f=F_2)
Area of the brakes = ![A_2=1.8 inches^2](https://tex.z-dn.net/?f=A_2%3D1.8%20inches%5E2)
Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.
![\frac{F_1}{A_2}=\frac{F_2}{A_2}](https://tex.z-dn.net/?f=%5Cfrac%7BF_1%7D%7BA_2%7D%3D%5Cfrac%7BF_2%7D%7BA_2%7D)
![F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb](https://tex.z-dn.net/?f=F_2%3D%5Cfrac%7B5.6%20lb%5Ctimes%201.8%20inhes%5E2%7D%7B0.6%20inches%5E2%7D%3D16.8%20lb)
16.8 lb is the force on the brake pad of one wheel.