The coefficient of static friction between the puck and the surface.
In fact, that coefficient describes exactly how "hard" it is to cause the puck to start moving, if it starts from an idle condition.
Let's say the velocity at the bottom of the window was "v."
s = v*t + ½at²
2 m = v * 1.3s - 4.9m/s² * (1.3s)² = v * 1.3s - 8.3 m
v = 10.3m / 1.3s = 7.9 m/s
Then the initial speed was
V = √(v² + 2as) = √(7.9m/s² + 2 * 9.8m/s² * 7.5m) = 14 m/s ◄ initial velocity
(after rounding to 2 digits from 14.5 m/s).
According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.
Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²
We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m
Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
= </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
= 39.6 N/C
Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
