Yes. Electric field lines are the imaginary lines that represents strength & direction of electric field due to a charge. It can be modeled as lines.
In short, Your Answer would be "True"
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Answer:
I'm not sure
Explanation:
sorry I'm not smart that's why I have this app
Answer:
c = 4,444.44
Explanation:
You have the following expression for the acceleration of the projectile:
(1)
s: distance to the ground of the projectile
To find the value of the constant c you use the following formula:
(2)
vo: initial velocity = 0 m/s
v: final speed = 200 m/s
Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m
You replace the expression (1) into the expression (2):

You do the constant c in the last equation, then you replace the values of v, s and Δs:

Answer:
(a) 1.21 m/s² (b) 1.75 m/s²
Explanation:
The initial speed of the car, u = 17.8 m/s
Case 1.
Final speed of the car, v = 23.5 m/s
Time, t = 4.68-s
Acceleration = rate of change of velocity

Case 2.
Final speed of the car, v = 15.3 m/s

Hence, this is the required solution.
Water cycle, evaporation, condensation, and freezing