Answer:
Explanation:36.05 km
Given
First car travels
South
then turns and travels
east
Suppose south as negative y axis and east as positive x axis
So, 

Displacement is the shortest between initial and final point
Dispalcement
Displacement
Displacement
Magnitude 
Magnitude
Answer:
Explanation:
Let the magnitude of magnetic field be B .
flux passing through the coil's = area of coil x field x no of turns
Φ = 3.13 x 10⁻⁴ x B x 135 = 422.55 x 10⁻⁴ B .
emf induced = dΦ / dt , Φ is magnetic flux.
current i = dΦ /dt x 1/R
charge through the coil = ∫ i dt
= ∫ dΦ /dt x 1/R dt
= 1 / R ∫ dΦ
= Φ / R
Total resistance R = 61.1 + 44.4 = 105.5 ohm .
3.44 x 10⁻⁵ = 422.55 x 10⁻⁴ B / 105.5
B = 3.44 x 10⁻⁵ x 105.5 / 422.55 x 10⁻⁴
= .86 x 10⁻¹
= .086 T .
Answer:
a)
a = 2 [m/s^2]
b)
a = 1.6 [m/s^2]
c)
xt = 2100 [m]
Explanation:
In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.
a)
When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:

where:
Vf = final velocity = 40 [m/s]
Vi = initial velocity = 0 (starting from rest)
a = acceleration [m/s^2]
t = time = 20 [s]
40 = 0 + (a*20)
a = 2 [m/s^2]
The distance can be calculates as follows:

where:
x1 = distance [m]
40^2 = 0 + (2*2*x1)
x1 = 400 [m]
Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.
v = x2/t2
where:
x2 = distance [m]
t2 = 30 [s]
x2 = 40*30
x2 = 1200 [m]
b)
Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.

Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.
0 = 40 - (a *25)
a = 40/25
a = 1.6 [m/s^2]
The distance can be calculates as follows:

0 = (40^2) - (2*1.6*x3)
x3 = 500 [m]
c)
Now we sum all the distances calculated:
xt = x1 + x2 + x3
xt = 400 + 1200 + 500
xt = 2100 [m]
The AMOUNT of energy the ball has doesn't change. It's 294 joules in Darwin's hand, and it's still 294 joules when the ball hits the ground. It's all PE before he let's it go, and it steadily changes from PE to KE all the way down.
It BEGINS to turn into KE immediately, when Darwin lets go of the ball, and it starts to fall.
More and more PE turns into KE as the ball falls, all the way down.
When the ball hits the ground, it has no more PE left. All of its mechanical energy is then KE.