The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.
Given the data in the question;
Since the brick was initially at rest before it was dropped,
- Initial Velocity;
- Height from which it has dropped;
- Gravitational field strength;
Final speed of brick as it hits the ground; 
<h3>Velocity</h3>
velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:
Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.
To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.
Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.
Learn more about equations of motion: brainly.com/question/18486505
Well first of all, a planet doesn't have a semimajor axis, although it's orbit does.
In an orbit with a smaller semimajor axis, the planet moves faster, and its orbital period is shorter.
That's why the International Space Station circles the Earth in less time than the Moon does.
Force is transferred from the moving ball to the stationary ball.
Answer:
The ball fell 275.625 meters after 7.5 seconds
Explanation:
<u>Free fall
</u>
If an object is left on free air (no friction), it describes an accelerated motion in the vertical direction, powered exclusively by the acceleration of gravity. The formulas needed to compute the different magnitudes involved are
Where 
is the final speed of the object in free fall, assumed positive downwards, t is the time elapsed since the release and y is the vertical distance traveled by the object
The ball was dropped from a cliff. We need to calculate the vertical distance the ball went down in t=7.5 seconds. We'll use the formula
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 : 
