Answer:
Explanation:
Total momentum of the system before the collision
.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left
If v be the velocity of the stuck pucks
momentum after the collision = 2 v
Applying conservation of momentum
2 v = - .75
v = - .375 m /s
Let after the collision v be the velocity of .5 kg puck
total momentum after the collision
.5 v + 1.5 x .231 = .5v +.3465
Applying conservation of momentum law
.5 v +.3465 = - .75
v = - 2.193 m/s
2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.
Kinetic energy before the collision
= 2.25 + 1.6875
=3.9375 J
kinetic energy after the collision
= .04 + 1.2 =1.24 J
So kinetic energy is not conserved . Hence collision is not elastic.
3 ) Change in the momentum of .5 kg
1.5 - (-1.0965 )
= 2.5965
Average force applied = change in momentum / time
= 2.5965 / 25 x 10⁻³
= 103.86 N
Yes, it can change states of matter from heat. Like ice, ice changes it’s state of matter from the temperature it is in, which causes it to turn into a liquid.(another form of matter) Again with heat, if water is boiled to a certain temperature it converts to a gas. Now with cold. If water is at such a coke temperature it will turn to ice, a form of matter. Now when a gas is cooled down, the atoms comprising the gas have less energy to move around.
Answer:
6010.457N
Explanation:
Centripetal acceleration = a= V²/R
At a radius of 3.6m and velocity of 16.12m/s,
Acceleration is
a = 16.12²/ 3.6 = 72.182 m/s²
Force = Mass (m) * Acceleration (a)
36 = m * 72.182
m = 36/72.182
At breaking point
Radius = 0.468 m and Velocity = 75.1 m/s
a = V²/R = 75.1²/0.468
a = 12051.3 m/s
F = Mass(m) * Acceleration (a)
F = m * 12051.3
m = F/ 12051.3
Settings the ratio of mass equal
m = m
=> 36/72.182 = F/12051.3
F = 12051.3 * 36/72.182
F = 6010.457N
15 min
Explanation:
take 0.25 and put it in for 1.00 and you will see its 0.25 but when you add it all 4 times it is 1.00 so then you would take that and do it to the hour ... how many times does four go into 60
You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x).
