Answer:
Second Trial satisfy principle of conservation of momentum
Explanation:
Given mass of ball A and ball B 
Let mass of ball
and
Final velocity of ball 
Final velocity of ball 
initial velocity of ball 
Initial velocity of ball 
Momentum after collision 
Momentum before collision 
Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.
Now, 
Plugging each trial in this equation we get,
First Trial

momentum before collision
moment after collision
Second Trial

moment before collision
moment after collision
Third Trial

momentum before collision
moment after collision
Fourth Trial

momentum before collision
moment after collision
We can see only Trial- 2 shows the conservation of momentum in a closed system.
Definitely ball and basket
False because faster moving slope would be going up and slower would be going down because its decreasing
Answer:
v₂ = 97.4 m / s
Explanation:
Let's write the Bernoulli equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Index 1 is for tank and index 2 for exit
We can calculate the pressure in the tank with the equation
P = F / A
Where the area of a circle is
A = π r²
E radius is half the diameter
r = d / 2
A = π d² / 4
We replace
P = F 4 / π d²2
P₁ = 397 4 /π 0.058²
P₁ = 1.50 10⁵ Pa
The water velocity in the tank is zero because it is at rest (v1 = 0)
The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa
Since the pipe is horizontal y₁ = y₂
We replace on the first occasion
P₁ = P₂ + ½ ρ v₂²
v₂ = √ (P1-P2) 2 / ρ
v₂ = √ [(1.50-1.013) 10⁵ 2/1000]
v₂ = 97.4 m / s
Answer:
The tension in the strap is 74.82 N.
Explanation:
Given that,
Angle between the horizontal and the suitcase is 36 degrees.
The distance traveled by the suitcase is 15 meters.
Let the work done by the suitcase is 908 J. We know that the work done in the vector form is given by :

So, the tension in the strap is 74.82 N. Hence, this is the required solution.