Answer:
The position of the particle is -2.34 m.
Explanation:
Hi there!
The equation of position of a particle moving in a straight line with constant acceleration is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the particle at a time t:
x0 = initial position.
v0 = initial velocity.
t = time
a = acceleration
We have the following information:
x0 = 0.270 m
v0 = 0.140 m/s
a = -0.320 m/s²
t = 4.50 s (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).
Then, we have all the needed data to calculate the position of the particle:
x = x0 + v0 · t + 1/2 · a · t²
x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²
x = -2.34 m
The position of the particle is -2.34 m.
Option(a) the mass of cart 2 is twice that of the mass of cart 1 is the right answer.
The mass of cart 2 is twice that of the mass of cart 1 is correct about the mass of cart 2.
Let's demonstrate the issue using variables:
Let,
m1=mass of cart 1
m2=mass of cart 2
v1 = velocity of cart 1 before collision
v2 = velocity of cart 2 before collision
v' = velocity of the carts after collision
Using the conservation of momentum for perfectly inelastic collisions:
m1v1 + m2v2 = (m1 + m2)v'
v2 = 0 because it is stationary
v' = 1/3*v1
m1v1 = (m1+m2)(1/3)(v1)
m1 = 1/3*m1 + 1/3*m2
1/3*m2 = m1 - 1/3*m1
1/3*m2 = 2/3*m1
m2 = 2m1
From this we can conclude that the mass of cart 2 is twice that of the mass of cart 1.
To learn more about inelastic collision visit:
brainly.com/question/14521843
#SPJ4
Answer:
Action and reaction are equal but act in opposite directions
Answer:
The answer is 13 however make sure if they ask for a certain measurement like meter answer it by saying 13 meters.
Explanation:
This basically turns into basic algebra if you know the formula for work. The formula for work is W=F*d
Here are the variables that you know 650J=50N*d so you need d.
All you do is divide 650J by 50N and you get a total of 13 (meters since I don't know what they want you to put it in).
