Answer:
acceleration 8 km/h/s south
Explanation:
First of all, let's remind that a vector quantity is a quantity which has both a magnitude and a direction.
Based on this definition, we can already rule out the following two choices:
distance: 40 km
speed: 40 km/h
Since they only have magnitude, they are not vectors.
Then, the following option:
velocity: 5 km/h north
is wrong, because the car is moving south, not north.
So, the correct choice is
acceleration 8 km/h/s south
In fact, the acceleration can be calculated as

where
v = 40 km/h is the final velocity
u = 0 is the initial velocity
t = 5 s is the time
Substituting,

And since the sign is positive, the direction is the same as the velocity (south).
Answer:
"Narrow the focus of research question"
Explanation:
O Narrow the focus of research question
This is good! You can still use your question, but focus in on something so you have a proper research project.
O Add another research question
Would adding another question to an already broad question help? No.
O Use the very first source you find for your project
If your question is too broad, you should not use whatever you see first as it may be incorrect or does not answer the question
O Change the scope of your project
You could, but if you have a set scope for your project (a) you might not be able to change it (b) you don't need to restart
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
Vf =Vi + at
Vi =Vf - at
= 34 - 15•13
= - 161 m/s :/
Answer:
It is the depth of the water that allowed it to be like day
Explanation:
what we see abovewater is not the actual thing or ssize we would see when insife the water
Answer:
a) 0.64 b) 2.17m/s^2 c) 8.668joules
Explanation:
The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,
Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move
Frictional force = mgsin20o + 5N = 6.71+5N = 11.71
The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44
Coefficient of static friction = 11.71/18.44= 0.64
Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)
b) coefficient of kinetic friction = frictional force/ normal force
Fr = 0.4* mgcos 20o = 7.375N
F due to motion = ma = total force - frictional force
Ma = 11.71 - 7.375 = 4.335
a= 4.335/2(mass of the block) = 2.17m/s^2
C) work done = net force *distance = 4.335*2= 8.67Joules