Answer: a) 6.67cm/s b) 1/2
Explanation:
According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".
Let m1 and m2 be the masses of the bodies
u1 and u2 be their velocities respectively
m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s
Since momentum = mass × velocity
The conservation of momentum of the body will be
m1u1 + m2u2 = (m1+m2)v
Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.
5(20) + 10(0) = (5+10)v
100 + 0 = 15v
v = 100/15
v = 6.67cm/s
Therefore the velocity of each object after the collision is 6.67cm/s
b) kinectic energy of the 10.0g object will be 1/2MV²
= 1/2×10×6.67²
= 222.44Joules
kinectic energy of the 5.0g object will be 1/2MV²
= 1/2×5×6.67²
= 222.44Joules
= 111.22Joules
Fraction of the initial kinetic transferred to the 10g object will be
111.22/222.44
= 1/2
Answer:
False.
Explanation:
Backbone Cabling: A system of cabling that connects the equipment rooms and telecommunications rooms.
Horizontal Cabling: The system of cabling that connects telecommunications rooms to individual outlets or work areas on the floor.
Answer: Pieces of minerals, rocks, plant and animal remains.
Explanation: Pieces of minerals, rocks, plant and animal remains. whether or not patterns cause flow rates of rivers to vary. sand settles from faster-moving water;smaller costs of silt and clay that form up mud settle from slower moving water.
Answer:
JA
Explanation:
s of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position. Assume all variable and constants are in SI units.
Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
