Answer:
Flood Plain
Explanation:
The amount of water that circulates through a river, the flow, varies in time and space. These variations define the hydrological regime of a river. Temporary variations occur during or just after episodes of rains or thaws. Much of the water that falls in the catchment basin circulates underground, or feeds underground aquifers and takes much longer to feed the river flow and can reach it days, weeks or months after the rain generated by the runoff. The runoff that goes to the river is what increases its flow. In extreme cases, flooding can occur when the water supply is greater than the river's ability to evacuate it, overflowing and covering nearby flat areas or floodplain. In this distribution between the runoff water (or stream) that goes directly to the channel and water that infiltrates, feeds the aquifers and maintains the flow in the river in times without precipitation depends largely on the geomorphological integrity of the entire river system .
In natural dynamics, the river systems have their own space that has been modeled by the floodwaters and is made up of the channel, the banks and the plain or flood plain. Its dimensions have been defined by the main flood events that this river has attended. Floodplains are wide and flat areas built by the river in its floodwaters. They are flooded frequently and are covered by sediments and nutrients that fertilize the soil act as natural reservoirs, reducing the speed of the downstream current. They store floodwater and rainfall in aquifers (underground area).
From the front door to his current position, Aaron's Distance is 4m. and his Displacement is 4m.
Distance is a measure of how far apart two objects or locations are using numbers. The distance can refer to a physical length in physics or to an estimate based on other factors in common usage. Sometimes, the symbol |AB| is used to represent the distance between two points.
A displacement is a vector in geometry and mechanics that have a length equal to the shortest distance between a point P's initial and final positions.
Given,
Length of path = 4m
Distance =?
Displacement =?
Distance is the length of the path and it has magnitude but no direction. The distance covered is 4m Displacement is the length of the path and it has magnitude and direction. Displacement is 4m towards the mailbox.
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Answer:
u make no sense can u just show a picture
Explanation:
Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.
<h3>Kinetic energy</h3>
Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.
Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.
The kinetic energy is represented by the following expression:
Ec= ½ mv²
Where:
- Ec is the kinetic energy, which is measured in Joules (J).
- m is the mass measured in kilograms (kg).
- v is the speed measured in meters over seconds (m/s).
<h3>Kinetic energy of a bullet</h3>
In this case, you know:
Replacing in the definition of kinetic energy:
Ec= ½ ×0.500 kg× (25 m/s)²
Solving:
<u><em>Ec= 156.25 J</em></u>
Finally, the bullet has a kinetic energy of 156.25 J.
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