Answer:
The sum of momenta of each object before and after collision is equal . Thus when they collide , their momentum will change but its sum will be sane as the sum of the momentum before collision
Answer:
Explanation:
graph would be a straight line from (0, 0) to (400, 8)
Plot points are
PE = mgh
50(0) = 0 J
50(2) = 100 J
50(4) = 200 J
50(6) = 300 J
50(8) = 400 J
Answer:
The answer is compound
Explanation:
heterogeneous mixture is wronggggg
Explanation:
The gravitational force equation is the following:

Where:
G = Gravitational constant = 
m1 & m2 = the mass of two related objects
r = distance between the two related objects
The problem gives you everything you need to plug into the formula, except for the gravitational constant. Let me know if you need further clarification.
Answer:
The tank is losing

Explanation:
According to the Bernoulli’s equation:
We are being informed that both the tank and the hole is being exposed to air :
∴ P₁ = P₂
Also as the tank is voluminous ; we take the initial volume
≅ 0 ;
then
can be determined as:![\sqrt{[2g (h_1- h_2)]](https://tex.z-dn.net/?f=%5Csqrt%7B%5B2g%20%28h_1-%20h_2%29%5D)
h₁ = 5 + 15 = 20 m;
h₂ = 15 m
![v_2 = \sqrt{[2*9.81*(20 - 15)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%2820%20-%2015%29%5D)
![v_2 = \sqrt{[2*9.81*(5)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%285%29%5D)
as it leaves the hole at the base.
radius r = d/2 = 4/2 = 2.0 mm
(a) From the law of continuity; its equation can be expressed as:
J = 
J = πr²
J =
J =
b)
How fast is the water from the hole moving just as it reaches the ground?
In order to determine that; we use the relation of the velocity from the equation of motion which says:
v² = u² + 2gh
₂
v² = 9.9² + 2×9.81×15
v² = 392.31
The velocity of how fast the water from the hole is moving just as it reaches the ground is : 
