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:
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Answer:
<em>The coefficient of static friction between the crate and the floor is 0.41</em>
Explanation:
<u>Friction Force</u>
When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.
The friction force when an object is moving on a horizontal surface is calculated by:
[1]
Where 
is the coefficient of static or kinetics friction and N is the normal force.
If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:
N = W = m.g
The crate of m=20 Kg has a weight of:
W = 20*9.8
W = 196 N
The normal force is also N=196 N
We can find the coefficient of static friction by solving [1] for :
The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:
The coefficient of static friction between the crate and the floor is 0.41
Answer:
find the sum of the inital and final velocitys and divide by 2 to find the average
Answer:
Solving for time :
(There are 4 formulas from linear motion. These formulas are very helpful as it allows us to prevent complicated calculations. Choose among the four that has : 1. The most constants known
2. The unknown constant that we want to solve)
s = (1/2)(u+v)t <--- one of the formulas
from linear motion
s (distance) = 0.05m
u (initial velocity) = 100m/s
v (final velocity) = 0 m/s (it stops)
t (time taken for change in velocity) = to be found
0.05 = (1/2)(100+0)t
t = 0.001 seconds
Solving for the resistant force :
Since the bullet hits the bag with an impulsive force and stops, the force that stops the bullet is the resistant force.
When the bullet stops :
F net = 0
F r = F imp
F r = (mu -mv)/t
F r = (0.01x100-0.01x0)/0.001
F r = 1/0.001
F r = 1000N
Answer:
Heat of vaporization will be 22.59 j
Explanation:
We have given mass m = 10 gram
And heat of vaporization L = 2.259 J/gram
We have to find the heat required to vaporize 10 gram mass
We know that heat of vaporization is given by 
, here m is mass and L is latent heat of vaporization.
So heat of vaporization Q will be = 10×2.259 = 22.59 J
