Answer:
3500j
0.250 kg
14.14 m/s
220500 j
93.91 m/s
Explanation:
K.E =1/2 m v^2
first question
1/2×700×10=3500 j
second question
2×78.2/25^2= 0.250 kg
Third question
8kj=8000j
root (2×8000/80)= 14.14 m/s
Last one
P.E=MGH
so it will be
P.E=50×450×9.8=220500 j
the speed
root (2×220500/50)=93.91 m/s
Current I is the rate at which charge moves through an area A, such as the cross-section of a wire
Direction of prevailing winds. shape of the land (known as 'relief' or 'topography') distance from the equator. the El Niño phenomenon.
To solve this problem we will apply the concept of Electric Flow, which is understood as the product between the Area and the electric field. For the data defined by the area, we will use the geometric measurement of the area in a circle (By the characteristics of the object) This area will be equivalent to,



Applying the concept of electric flow we have to

Replacing,


Therefore the magnitude of the electric field is 
Explanation:
Below is an attachment containing the solution.