First we will use the concepts of motion kinetics for which the final speed is defined as shown below,

Here,
= Final velocity
= Initial velocity
a = Acceleration
s = Distance
Replacing,


Using the conservation of energy for kinetic energy we have,



Therefore the kinetic energy of the car is 31900J
1- interaction between 2 objects
2- action- reaction force pairs
Answer:
x = 1474.9 [m]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.
We must understand that when forces are applied on the body, they tend to slow the body down to stop it.
So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.
ΣF = m*a
![10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]](https://tex.z-dn.net/?f=10%20%2B12%2Asin%2860%29%3D%20-%206%2Aa%5C%5Ca%20%3D%20-%203.39%5Bm%2Fs%5E%7B2%7D%5D)
Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

where:
Vf = final velocity = 0 (the block stops)
Vo = initial velocity = 100 [m/s]
a = - 3.39 [m/s²]
x = displacement [m]
![0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]](https://tex.z-dn.net/?f=0%20%3D%20100%5E%7B2%7D-2%2A3.39%2Ax%5C%5Cx%3D%5Cfrac%7B10000%7D%7B2%2A3.39%7D%5C%5Cx%3D1474.9%5Bm%5D)
Answer:
3 m/s squared
Explanation:
The formula you use is Vf= Vi + at. You rearrange it to a= Vf - Vi/t. The Vf is 27m/s. The Vi is 0m/s and the t is 9s. Cross out Vi since it’s zero and you’re left with a= 27m/s divided by 9s, which equals 3
Answer: y = 2.4×10^-6m or y= 2.4μm
Explanation: The formulae for the distance between the central bright fringe to any other fringe in pattern is given as
y = R×mλ/d
Where y = distance between nth fringe and Central bright spot fringe.
m = position of fringe = 4
λ = wavelength of light= 600nm = 600×10^-9 m
d = distance between slits = 1.50×10^-5m
R = distance between slit and screen = 2m
y = 2 × 4 × 600×10^-9/2
y = 4800×10^-9/2
y = 2400 × 10^-9
y = 2.4×10^-6m or y= 2.4μm