Answer:
a) 
b) the motorcycle travels 155 m
Explanation:
Let
, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

where:
is the speed of the motorcycle at time 2
is the velocity of the car (constant)
is the velocity of the car and the motorcycle at time 1
d is the distance between the car and the motorcycle at time 1
x is the distance traveled by the car between time 1 and time 2
Solving the system of equations:
![\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dcar%26motorcycle%5C%5Cx%3Dv_0%5CDelta%7Bt%7D%26x%2Bd%3D%28%5Cfrac%7Bv_0%2Bv_%7Bm2%7D%7D%7B2%7D%7D%29%20%5CDelta%7Bt%7D%5Cend%7Barray%7D%5Cright%5D)

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

Answer:
No, distance is more important.
The acceleration of the car would be 0.33 first and then it would be 0.17.
<u>Explanation:</u>
An applied force is a force that is applied to an object by an individual or another item. On the off chance that an individual is pushing a work area over the room, at that point there is an applied power following up on the article. The applied power is the power applied on the work area by the individual.
The net force applied to the object rises to the mass of the article increased by the measure of its acceleration. The net power following up on the soccer ball is equivalent to the mass of the soccer ball duplicated by its adjustment in speed each second (its acceleration).
Answer:
pretty sure for this one, "if the density affected the liquid's ability to retain heat."
Explanation:
a hypothesis should be in "if, then, because.." format, and while this bit of the passage doesnt include all three, it does include one! that bit of the passage would be an incomplete hypothesis.
Explanation:
Given that,
The optical power of the equivalent single lens is 45.4 diopters.
(a) The relationship between the focal length and the focal length is given by:


f = 0.022 m
or
f = 2.2 cm
(b) We need to find how far in front of the retina is this "equivalent lens" located. It is given by using lens formula as :

Here, u = infinity

v = 2.2 cm
So, at 2.2 cm in front of the retina is this "equivalent lens" located.
Hence, this is the required solution.