5 km northeast. Left and up would make northeast
Answer:

Explanation:
Given two mass on an incline code
and
and an angle of inclination
.
. Assume that
is the weight being pulled up and
the hanging weight.
-The equations of motion from Newton's Second Law are:
where a is the acceleration.
#Substituting for
(tension) gives:

#and solving for 
which is the system's acceleration.
Initial speed = 2√10 m/s
<h3>Further explanation </h3>
Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration
An equation of uniformly accelerated motion
V = vo + at
Vt² = vo² + 2a (x-xo)
x = distance on t
vo / vi = initial speed
vt / vf = speed on t / final speed
a = acceleration
vf=20 m/s
d = 60 m
a = 3 m/s²

Answer:
DONTKNOW NEED CAREER FILD FIRST TO ANWSER
Explanation:
Answer:

Explanation:
Δ
- Δ
is the difference in velocity before and after a given time.
is the acceleration of the object during this time.
is time
is another way to write this equation.
- The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δ


- I rearranged this equation to solve for
, but this is a step that you don't need to take, it's just good to get in the habit of doing this. - Plug in the given values. Note that our final velocity is
, because the car travels until at <em>rest</em>.
![a=\frac{v_f-v_i}{t}\\a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv_f-v_i%7D%7Bt%7D%5C%5Ca%3D%5Cfrac%7B%280%29-%5B%2817.1%5Cfrac%7Bmiles%7D%7Bhour%7D%20%29%28%5Cfrac%7Bhour%7D%7B3600s%7D%29%28%5Cfrac%7B1609.34m%7D%7Bmile%7D%29%5D%7D%7B9.7s%7D)
- Our initial velocity is in mph, something not in standard units, so if not changed, you will get an incorrect answer. What you need to do is cancel out the units your prior value had using division and multiplication, and at the same time multiply and divide the correct numbers and units into your equation. Or look up a converter.
![a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}\\a=\frac{0m/s-7.6m/s}{9.7s} \\a=\frac{-7.6m/s}{9.7s}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%280%29-%5B%2817.1%5Cfrac%7Bmiles%7D%7Bhour%7D%20%29%28%5Cfrac%7Bhour%7D%7B3600s%7D%29%28%5Cfrac%7B1609.34m%7D%7Bmile%7D%29%5D%7D%7B9.7s%7D%5C%5Ca%3D%5Cfrac%7B0m%2Fs-7.6m%2Fs%7D%7B9.7s%7D%20%5C%5Ca%3D%5Cfrac%7B-7.6m%2Fs%7D%7B9.7s%7D)
- if you converted correctly, your answer for
will be ≅
. - Now divide. Notice that the units for acceleration are
or <em>meters per second, per second</em>.

- Our final answer is <em>negative </em>because the car is <em>slowing down</em>. Do not square this answer as the square symbol only applies to the units, not the magnitude.