C.
The force of friction = coefficient of friction * normal force.
Adding the book to the boy increases the normal force and the component of the gravitational force directed down the slide. This in turn increases the force of friction as can be seen by the relationship from the above equation. For a stationary object, the force of static friction is always equal to the force applied (in this case, it is the component of the gravitational force directed down the slide). That means that so long as the boy is not moving and his mass increases, the frictional force is increasing also to balance the increased downward gravitation force directed down the slide.
Chemical composition, the element no longer retains the same chemical formula as it did before.
Answer:
Vx = 35.31 [km/h]
Vy = 18.77 [km/h]
Explanation:
In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.
![v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h]](https://tex.z-dn.net/?f=v_%7Bx%7D%20%3D%2040%2Acos%2828%29%5C%5CV_%7Bx%7D%20%3D%2035.31%20%5Bkm%2Fh%5D)
In order to find the vertical component, we must use the sine function of the angle.
![V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h]](https://tex.z-dn.net/?f=V_%7By%7D%3D40%2Asin%2828%29%5C%5CV_%7By%7D%20%3D%2018.77%20%5Bkm%2Fh%5D)
Answer:
C
Explanation:
Ur welcome just took on test
Answer:
t = 0.33h = 1200s
x = 18.33 km
Explanation:
If the origin of coordinates is at the second car, you can write the following equations for both cars:
car 1:
(1)
xo = 10 km
v1 = 55km/h
car 2:
(2)
v2 = 85km/h
For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:


The position in which both cars coincides is:
