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
<h2>
Answer:</h2>
(a) 10N
<h2>
Explanation:</h2>
The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force 
is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F -
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F - = m x a
F - = 50 x 0
F - = 0
F = -------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force is opposing the movement of the box.
∑F = 1.5F -
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F - = m x a
1.5F - = 50 x 0.1
1.5F - = 5 ---------------------(iii)
<em>Substitute </em><em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
<em />
ThIs is the same type of problem
find out the time value
3 = 1/2*a*T^2
6/10 = t^2
t = 0.77 seconds
and the distance is given 5 m
thus speed ,= distance/time
speed = 5/0.77
= 6.45 m/s
Answer:
vDP = 21.7454 m/s
θ = 200.3693°
Explanation:
Given
vDE = 7.5 m/s
vPE = 20.2 m/s
Required: vDP
Assume that
vDE to be in direction of - j
vPE to be in direction of i
According to relative motion concept the velocity vDP is given by
vDP = vDE - vPE (I)
Substitute in (I) to get that
vDP = - 7.5 j - 20.2 i
The magnitude of vDP is given by
vDP = √((- 7.5)²+(- 20.2)²) m/s = 21.7454 m/s
θ = Arctan (- 7.5/- 20.2) = 20.3693°
θ is in 3rd quadrant so add 180°
θ = 20.3693° + 180° = 200.3693°
Answer:
L/D= 112
Explanation:
Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.
Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.
Lift increases proportionally to the square of the speed.
The solutions to the question is the file attached to this explanation.
Lift,L= qC(l). S---------------------------(1).
and,
Drag,D = qC(d).S ----------------------(2).
Hence, Lift to drag ratio,L/D= C(l)/C(d).
Therefore, we have to compute various angle of attack.(check attached file)...
Then, (L/D) will then be equal to 112.
