Answer:
U2 = 47.38m/s = initial velocity of B before impact
Explanation:
An example of the diagram is shown in the attached file because of missing angle of direction in the question
Mass A, B are mass of cars
A = 1965
B =1245
U1 = initial velocity of A = 52km/hr
U2 = initial velocity of B
V = common final velocity of two cars
BU2 = (A + B)*V sin ¤ ...eq1 y plane
AU1 = (A + B) *V cos ¤ ....equ 2plane
From equ 2
V = AU1/(A + B)*cos ¤
Substitute V into equation 1
We have
U2 = (AU1/B)tan ¤ where ¤ = angle of direction which is taken to be 30°
Substitute all parameters to get
U2 = (1965/1245)*52 * tan 30°
U2 = 47.38m/s
The answers is 30 miles per hour, the driver is speeding the car up, section-H, 12 minutes, section-D, and 65 miles per hour.
I think it's C, longer wave length.
The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration <em>a</em> of
950 N = (2300 kg) <em>a</em>
<em>a</em> = (950 N) / (2300 kg)
<em>a</em> ≈ 0.413 m/s²
Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of
<em>v</em> = -5.5 m/s + <em>a</em> (11.5 s)
<em>v</em> = -0.75 m/s
which means the wind slows the boat down to a velocity of 0.75 m/s westward.
