Answer: its speed upon release is 26.05 m/s
Explanation:
Given that;
mass m = 0.244 kg
force F = 30.3 N
V1 = 14.7 m/s
r = 59.3 cm = 0.593 m
Vf = ?
we know that;
1/2mV1² + FπR = 1/2mVf²
so we substitute
[1/2×0.244×(14.7)²] + [30.3×π×0.593 = 1/2×0.244×Vf²
26.3629 + 56.4478 = 0.122Vf²
82.8107 = 0.122Vf²
Vf² = 82.8107 / 0.122Vf
Vf² = 678.7762
Vf = √678.7762
Vf = 26.05 m/s
Therefore its speed upon release is 26.05 m/s
There's nothing on the list you provided that's related to Coulomb's Law.
Answer:
Explanation:
Ignoring air resistance
Initial vertical velocity is 30sin35 = 17.2 m/s
Gravity reduces this velocity to zero in a time of
t = v/g =17.2 / 9.8 = 1.755 s
it takes the same time to come back down to ground level for a total flight time of 2(1.755) = 3.51 s
The horizontal velocity is 30cos35 = 24.57 m/s
the distance traveled horizontally is
d = vt = 24.57(3.51) = 86.298... = 86 m
Answer:
This is the answer: The speed of a proton is about 5.0 × 10⁵ m/s
Explanation:
Because of the speeds of protons! :D
Answer:
The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
Explanation:
Given that
q₁ = 5 μ C
q₂ = - 4 μ C
The distance between charges = 50 cm
d= 50 cm
Lets take at distance x from the charge μ C ,the electrical field is zero.
That is why the distance from the charge - 4 μ C = 50 - x cm
We know that ,electric field is given as
Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.