The kinetic energy of the small ball before the collision is
KE = (1/2) (mass) (speed)²
= (1/2) (2 kg) (1.5 m/s)
= (1 kg) (2.25 m²/s²)
= 2.25 joules.
Now is a good time to review the Law of Conservation of Energy:
Energy is never created or destroyed.
If it seems that some energy disappeared,
it actually had to go somewhere.
And if it seems like some energy magically appeared,
it actually had to come from somewhere.
The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision. The large ball
and the small ball will just have to share the same 2.25 joules.
Answer:
The frictional force is 
Explanation:
From the question we are told that
The coefficient of kinetic force is μk = 0.35
The normal force felt by the puck is 
Generally the frictional force that acts on the puck is mathematically represented as
=> 
=>
Answer:
(1) it is transparent so it makes the reading difficult. (2) it is volatile. (3) it is a poor conductor of heat. (4) it has a higher specific heat capacity, so it absorbs more heat from the body with which it is kept in contact. (5) Water cannot be used in thermometer because of its higher freezing point and lower boiling point than other liquids . If water is used in a thermometer , it will start phase change at 0oC and 100oC and will not measure temperature , out of this range .
PLEASE READ CAREFULLY AND PLEASE MARK AS BRAINLIEST :)
Answer:
m1/m2 = 0.51
Explanation:
First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:
V = √F/u
This is the equation that describes the relation between speed of a pulse and a force exerted on it.
the value of "u" is:
u = m/L
Where m is the mass of the rod, and L the length.
Now, for the rod 1:
V1 = √F/u1 (1)
rod 2:
V2 = √F/u2 (2)
Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:
1.4V2 = √F/u1 (3)
Replacing (2) in (3):
1.4(√F/u2) = √F/u1 (4)
Now, let's solve the equation 4:
[1.4(√F/u2)]² = F/u1
1.96(F/u2) =F/u1
1.96F = F*u2/u1
1.96 = u2/u1 (5)
Now, replacing the expression of u into (5) we have the following:
1.96 = m2/L / m1/L
1.96 = m2/m1 (6)
But we need m1/m2 so:
1.96m1 = m2
m1/m2 = 1/1.96
m1/m2 = 0.51
