The viscous force on an object moving through air is proportional to its velocity.
The only forces acting on an object when falling are air resistance and its weight itself. The weight acts vertically downwards whereas air resistance acts vertically upward.
Let F be the viscous force due to air molecules, B be buoyant force due to air and W be the weight of falling object. Initially, the velocity of falling object and hence the viscous force F is zero and the object is accelerated due to force
(W-B). Because of the acceleration the velocity increases and accordingly the viscous force also increases. At a certain instant, the viscous force becomes equal to W-B. The net force then becomes zero and the object falls with constant velocity. This constant velocity is called terminal velocity.
Thus at terminal velocity, air resistance and force of gravity becomes equal.
Answer:
Hi... Potential energy is converted to kinetic energy and kinetic energy is converted to potential energy
Answer:
The water level rises more when the cube is located above the raft before submerging.
Explanation:
These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.
Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.
When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.
Vc = 0.45*0.45*0.45 = 0.0911 [m^3]
Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.
![Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]](https://tex.z-dn.net/?f=Ro_%7BH2O%7D%3DR_%7Bc%2Br%7D%5C%5Cwhere%3A%5C%5CRo_%7BH2O%7D%3D%20water%20density%20%3D%201000%20%5Bkg%2Fm%5E3%5D%5C%5CRo_%7Bc%2Br%7D%3D%20combined%20density%20cube%20%2B%20raft%20%5Bkg%2Fm%5E3%5D)
Density is given by:
Ro = m/V
where:
m= mass [kg]
V = volume [m^3]
The buoyancy force can be calculated using the following equation:
![F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]](https://tex.z-dn.net/?f=F_%7BB%7D%3DW%3DRo_%7BH20%7D%2Ag%2AVs%5C%5CW%20%3D%20%28200%2B730%29%2A9.81%5C%5CW%3D9123.3%5BN%5D%5C%5C%5C%5C9123%3D1000%2A9.81%2AVs%5C%5CVs%20%3D%200.93%20%5Bm%5E3%5D)
Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.
Answer: 
Explanation:
The diffraction angles
when we have a slit divided into
parts are obtained by the following equation:
(1)
Where:
is the width of the slit
is the wavelength of the light
is an integer different from zero.
Now, the second-order diffraction angle is given when
, hence equation (1) becomes:
(2)
Now we have to find the value of
:
(3)
Then:
(4)
(5)
Finally:
(6)
Work needed = 23,520 J
<h3>Further explanation</h3>
Given
height = 12 m
mass = 200 kg
Required
work needed by the crane
Solution
Work is the transfer of energy caused by the force acting on a moving object
Work is the product of force with the displacement of objects.
Can be formulated
W = F x d
W = Work, J, Nm
F = Force, N
d = distance, m
F = m x g
Input the value :
W = mgd
W = 200 kg x 9.8 m/s²x12 m
W = 23520 J