The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick
together and continue on with velocity vf. Which of these statements is true?
A. vf=v1
B. vf=v2
C. vf is less than v2
D. vf is less than v2 but greater than v1
E. vf>v1
A. vf=v1
B. vf=v2
C. vf is less than v2
D. vf is less than v2 but greater than v1
E. vf>v1
1 answer:
Answer:
See bolded below.
Explanation:
Consider the " Before " and " After. " " Before, " this particle 1 was trying to catch up with this particle 2, and " after " particle one had collided with particle two. Take a look at the attachment below for a more detailed examination.
Here is how this will play out. Particle 1, with great velocity, will hit particle 2, which would mean that Particle 2 has less velocity than Particle 1. Now after the collision, energy is transferred to Particle 2, and while Particle 1 has now stopped in it's tracks, Particle 2 - with more energy than before - will continue as long as it has to before friction eventually brings it to a stop.
_______________________________________________________
From this we can conclude that Vf, from the picture below, must have less energy than V1, but more energy than V2 - and vice versa.
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Answer:
It is unsafe
Explanation:
v = Velocity of car = 34 m/s
r = Radius of turn = 190 m
= Coefficient of static friction = 0.5
m = Mass of car = 1600 kg
g = Acceleration due to gravity = 9.81 m/s²
The centripetal force is given by
The frictional force is given by
If the centripetal force is greater than the frictional force then the car will slip which makes it unsafe.
Here, the centripetal force is greater than the frictional force which makes it unsafe to drive it at that speed.
Explanation:
The clowns need to leave the diving boards with enough horizontal velocity such that each travels 15 m (half the width of the pool) in the same time that they fall (vertically) the 10-m from the top of the diving board.
We'll assume no force acts on the clowns horizontally to slow them down while they are in flight. And we'll assume that only gravity acts on the clowns vertically.
We can treat the horizontal and vertical components separately. This will help simplify the problem.
Let's start with the vertical displacement. Let's say the clown is dropped from a height of 10-m. How long would it take them to fall that distance?
Using our equations of motion (with constant, linear acceleration), we can solve for this time.
Where d is the distance travelled, v is the initial velocity, a is acceleration, and t is time.
If the clown is dropped, they have an initial velocity of 0 (zero). We assumed only gravity acts on the clown, so acceleration equals the gravitational acceleration on earth.
The distance the clown travels is 10-m from the diving board to the surface of the water.
Let's solve.
Now that we know time, we can calculate how fast the clown needs to be running when they leap from the diving board to cover a distance of 15-m (Remember, half the width of the pool.)
Using our equations of motion, we know that
We assumed no forces act horizontally on the clown, therefore a = 0. We just need to solve for v. Substituting in the time we just solved for, we get something like this.
I'll leave it to you to solve this equation.