Answer:as per as Newtons second law, The forces exerted on the rope create tension.
As such,The tension is equal to the applied force.The tension is trasmitted to the opposite side and of the rope delivering the applied force.
Hope this helps.. :)
Answer:
Newton's second law states that when a body of mass m is accelerated with force f
then F=ma
this means acceleration of an object depends on both force with which it is moving as well as its mass
Answer:
53.13 °
Explanation:
In order to do this, we just need to apply the following:
tanα = Dy/Dx
Where:
Vy: speed of the ball in the y axis.
Vx: speed of the ball in the x axis.
At this point we do not need the speed of the first ball after the collision because in that moment is already heading in the direction that we are looking for. Therefore, we just need to use the innitial data to calculate the direction which the first ball will go.
According to this, then:
tanα = (40/30)
tanα = 1.3333
α = tan⁻¹(1.3333)
<h2>
α = 53.13°</h2>
This means that the final direction of the first ball is 53.13° and in the x axis because the starting momentum of this ball in the x axis has not dissapeared.
Hope this helps
Number 5 is A
number 6 is D
Answer:
1) 1.31 m/s2
2) 20.92 N
3) 8.53 m/s2
4) 1.76 m/s2
5) -8.53 m/s2
Explanation:
1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

So the maximum acceleration on the block is

4)As the box slides, it is now subjected to kinetic friction, which is

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is
28.25 / 16 = 1.76 m/s2
5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2