Given: distance 1 d₁ = 40 m; distance 2 d₂ = 3.8 m g = -9.8 m/s²
Initial Velocity Vi = 0 Final Velocity of stone 2 is unknown = ?
Total distance dₓ = d₁ - d₂ = 40 m - 3.8 m = 36.2 m
Formula: a = Vf² - Vi²/2d derive for Final Velocity Vf
acceleration is now due to gravity, therefore a = g
Vf = √2gd Vf = √2(9.8 m/s²)(36.2 m)
Vf = 26.64 m/s
Reason: The second stone will still start from rest.
Answer:
A) The particle will accelerate in the direction of point C.
Explanation:
As we know that
potential at points A, B,C and D as V_A, V_B, V_C, V_D and it is clear from the question that
V_A>V_B>V_C
And we know that flow is always from higher to lower potential (for positive charge due to positive potential energy).
So the charge will accelerate from B toward C.
Hence, the correct option is A.
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Answer:
Distance, d = 0.1 m
It is given that,
Initial velocity of meson,
Finally, the meson is coming to rest v = 0
Acceleration of the meson, (opposite to initial velocity)
Using third equation of motion as :
s is the distance the meson travelled before coming to rest.
So,
s = 0.1 m
The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.
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