According to Newton's first law, when the net force on the body is zero, the result will be that the body will have constant velocity. So when the force of friction on fireman's hands and the weight are equal, the fireman will slide down the pole with a constant velocity.
<span>it accelerates (or decelerates) depending on the magnitude of the net force and the direction of the net force with respect to the direction of motion.
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Answer:
Explanation:
Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each moving at the same speed v when they collide at the origin of the coordinates and stick together. After the collision, the masses move at an angle −θ2 with respect to the +x axis at speed v2 .1. What was the angle φ?
from the principle of momentum
In a system of colliding bodies,we know that the total momentum before collision will equal to the total momentum after collision.
Take note that momentum is the product of mass and velocity
momentum before collision=momentum after collision
mass, m
u=initial velocity of the identical masses
v2=the common velocity after the collision
Note that the collision is inelastic , since they both moved with the same velocity
umcosφ+umcosφ=(m+m)v2cos−θ2
2mucosφ=2mv2cos−θ2
To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:
Where,
x= Displacement
= Initial velocity
a = Acceleration
t = time
Since there is no initial velocity, the same equation can be transformed in terms of length and time as:
For the second cart
When the tenth car is aligned the length will be 9 times the initial therefore:
When the tenth car has passed the length will be 10 times the initial therefore:
The difference in time taken from the second car to pass it is 5 seconds, therefore:
From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:
From the relationship when the car has passed and the time difference we will have to:
Replacing the value found in the equation given for the second car equation we have to:
Finally we will have the time when the cars are aligned is
The time when you have passed it would be:
The difference between the two times would be:
Therefore the correct answer is C.
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