What are frames of reference
2 answers:
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Answer:
So the repulsive force between the pith ball will be
Explanation:
We have given that the pith ball have the equal charge q = -40 nC =
Distance between the charges = 10 cm =0.1 m
According to coulombs law
So the repulsive force between the pith ball will be
Answer:
A) t = 10.56 s, B) x = 235 m, C) v = 25.2 m / s
Explanation:
A) We can solve this problem using kinematics expressions.
The distance traveled by the truck is
x_c = v_c t
Distance traveled by the car.
The car must travel the distance that separates them from the truck x₀=25.0. Return to the lane at x₁ = 26.5 m. the length of the truck x₂=20.7m and the length of the car x₃ = 2 4.5 = 9 m, therefore the total length traveled by the car is
x_t = x₁ + x₂ + x₃
x_t = 26.5 + 20.7 +9 = 56.2 m
the distance traveled by the car when it returns to the lane is
x_c + x_t = x₀ + v₀ t + ½ a t²
when the car passes the car the distance traveled by the two vehicles is the same, we substitute
v_c t + x_t = x₀ + v₀ t + ½ a t²
½ a t² + t (v₀ -v_c) + (x₀ - x_t) = 0
we substitute the values
½ 0.560 t² + t (19.3 -19.3) + (25.0 - 56.2) =
0.28 t² -31.2 = 0
t =
t = 10.56 s
This is the time it takes for the car to pass the truck and back into the lane.
B) the distance traveled is
x = v₀ t + ½ a t²
x = 19.3 10.56 + ½ 0.560 10.56²
x = 235 m
C) the final velocity is
v = v₀ + a t
v = 19.3 + 0.560 10.56
v = 25.2 m / s