A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou
nd. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.
1 answer:
Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
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Answer:
a) V = -0.227 mV
b) V = -0.5169 mV
Explanation:
a)
Inside a sphere with a uniformly distributed charge density, electric field is radial and has a magnitude
E = (qr) / (4πε₀R³)
As we know that
V = -
By solving above equation, we get
V = (-qr²) / (8πε₀R³)
When
R = 1.81 cm
r = 1.2 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-2.80 × 10⁻¹⁵ × (1.2 × 10⁻²)²) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²)³)
V = -2.27 × 10⁻⁴ V
V = -0.227 mV
b)
When
r = R
R = 1.81 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-qR²) / (8πε₀R³)
V = (-q) / (8πε₀R)
V = (-2.80 × 10⁻¹⁵) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²))
V = -5.169 × 10⁻⁴ V
V = -0.5169 mV