As these are distances created by moving in a straight line, using a trigonometric analysis can solve the missing single straight-line displacement. Looking at the 48m and 12m movements as legs of a triangle, obtaining the hypotenuse using the pythagorean theorem will yield us the correct answer.
This is shown below:
c^2 = 48^2 + 12^2
c = sqrt(2304 + 144)
c = sqrt(2448)
c = 49.48 m
To obtain the angle at which Anthony walks 49.48, we obtain the arc tangent of (12/48). This is shown below:
arc tan (12/48) =14.04 degrees.
Therefore, Anthony could have walked 49.48 m towards the S 14.04 W direction.
They determine if it has a negative or positive charge
Answer:
Rmax = 3.4 10⁶ m
Explanation:
For this exercise we will use the concept of energy
Initial. On the surface of the luma
Em₀ = K + U
Em₀ = ½ m v² - G m M / R_moon
Final. At the furthest point
Emf = U
Emf = - g m M / R_max
Em₀ = Emf
½ m v² - G m M / R_moon = - G m M / R_max
½ v² + G M (-1 / R_moon + 1 / R_max) = 0 (1)
Let's use the fact that tells us that the speed of the rocket is equal to the speed of a satellite that rotates around the moon near the surface, let's use Newton's second law
F = m a
Acceleration is centripetal
a = v² / r
r= R_moon
G m M / R_moon² = m v² / R_mon
G M / R_moon = v²
We substitute in 1
½ G M / R_moon + G M (1 / R_max - 1 / R_moon) = 0
1 / R_max = 1 / R_moon (1- ½)
R_max = R_moon 2
Rmax = 2 1700 103
Rmax = 3.4 10⁶ m
Recall that
where 
and 
are initial and final velocities, respectively; 
is acceleration; and 
is the net displacement, or distance if the object is moving in a single direction.
The car has initial speed 20 m/s and acceleration -3.25 m/s². It comes to a stop, so it has 0 final speed. Then
0² - (20 m/s)² = 2 (-3.25 m/s²) ∆<em>x</em>
∆<em>x</em> = (20 m/s)² / (7.5 m/s²) ≈ 53.3 m
