Answer:
the electric field is 3.91 x 10⁶ N/C
Explanation:
Given the data in the question;
Electric field at a point due to point charge is;
E = kq/r²
where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator
Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C
so we substitute into the formula
E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²
E = 59400000 / 15.21
E = 3.91 x 10⁶ N/C
Therefore, the electric field is 3.91 x 10⁶ N/C
Answer:
18.032m/s, 16.59m
Explanation:
Used to find the distance traveled
y=1/2gt^2, where y= vertical distance, g=acceleration due to gravity, and t=time
y=1/2*9.8*1.84^2
y=16.59m distance
Used to solve the final velocity before hitting the ground
vf=vo+at where vf=final velocity, vo=initial velocity, a=acceleeration, t=time
vf=0+9.8*1.84
vf=18.032m/s
The carrying capacity of a biological species in an environment is the maximum population size of the species that the environment can sustain indefinitely, given the food, habitat, water, and other necessities available in the environment.
It increases. As it moves it <span>increases while the movement is in process.
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