Answer:
The potential difference between the plates is ![8.4\times10^{3}\ V](https://tex.z-dn.net/?f=8.4%5Ctimes10%5E%7B3%7D%5C%20V)
Explanation:
Given that,
Distance = 1.4 mm
Electric field strength ![E= 6.0\times10^{6}\ N/C](https://tex.z-dn.net/?f=E%3D%206.0%5Ctimes10%5E%7B6%7D%5C%20N%2FC)
Let the potential difference is V.
We need to calculate the potential difference between the plates
Using formula of electric field
![E=\dfrac{V}{d}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7BV%7D%7Bd%7D)
![V=Ed](https://tex.z-dn.net/?f=V%3DEd)
Where, V = potential
d = distance
Put the value into the formula
![V=6.0\times10^{6}\times1.4\times10^{-3}](https://tex.z-dn.net/?f=V%3D6.0%5Ctimes10%5E%7B6%7D%5Ctimes1.4%5Ctimes10%5E%7B-3%7D)
![V=8.4\times10^{3}\ V](https://tex.z-dn.net/?f=V%3D8.4%5Ctimes10%5E%7B3%7D%5C%20V)
Hence, The potential difference between the plates is ![8.4\times10^{3}\ V](https://tex.z-dn.net/?f=8.4%5Ctimes10%5E%7B3%7D%5C%20V)
The eroded rock and soil materials that are transported downstream by a river are called its load. A river transports, or carries, its load in three different ways: in solution, in suspension, and in its bed load.
Mineral matter that has been dissolved from bedrock is carried in solution. Common minerals carried in solution by rivers include dissolved calcium, magnesium, and bicarbonate. Most of a river’s solution load comes from groundwater seeping into the river. Before it reaches the stream,thegroundwaterhastraveledthroughfracturesinthebedrock, chemically eroding rock along the way.
When river water looks muddy, it is carrying rock material in suspension. Suspended material includes clay, silt, and fine sand. Although these suspended materials are heavier than water, the turbulence of the stream flow stirs them up and keeps them from sinking. Turbulence includes swirls and eddies that form in water as a result of friction between the stream and its channel. The faster a stream flows, the more turbulent and muddy it becomes. A rough or irregular channel also increases turbulence.
A river may also transport rock materials in its bed load. The bed load consists of sand, pebbles, and boulders that are too heavy to be carried in suspension. These heavier materials are moved along the streambed, especially during floods. Boulders and pebbles roll or slide along the river bed. Large sand grains are pushed along the bottom in a series of jumps and bounces.
The relative amounts of a river’s load that are carried in solution, in suspension, and in the bed load depend on the nature of the river, the climate, the type of bedrock, and the season of the year. As a general rule, most of the load carried by the world’s streams and rivers is carried in suspension. The size of a river’s suspended load increases with human land use. Road and building construction and removal of vegetation make it easier for rain to wash sediment into streams and rivers.
The value of 'g' is not affected by rotation at any place on Earth.
The hotter molecules become, the faster they move around. The colder they are, the more slow and lethargic they are
Answer:
l= 4 mi : width of the park
w= 1 mi : length of the park
Explanation:
Formula to find the area of the rectangle:
A= w*l Formula(1)
Where,
A is the area of the rectangle in mi²
w is the width of the rectangle in mi
l is the width of the rectangle in mi
Known data
A = 4 mi²
l = (w+3)mi Equation (1)
Problem development
We replace the data in the formula (1)
A= w*l
4 = w* (w+3)
4= w²+3w
w²+3w-4= 0
We factor the equation:
We look for two numbers whose sum is 3 and whose multiplication is -4
(w-1)(w+4) = 0 Equation (2)
The values of w for which the equation (2) is zero are:
w = 1 and w = -4
We take the positive value w = 1 because w is a dimension and cannot be negative.
w = 1 mi :width of the park
We replace w = 1 mi in the equation (1) to calculate the length of the park:
l= (w+3) mi
l= ( 1+3) mi
l= 4 mi