Answer:
Option B. 3.0×10¯¹¹ F.
Explanation:
The following data were obtained from the question:
Potential difference (V) = 100 V.
Charge (Q) = 3.0×10¯⁹ C.
Capacitance (C) =..?
The capacitance, C of a capacitor is simply defined as the ratio of charge, Q on either plates to the potential difference, V between them. Mathematically, it is expressed as:
Capacitance (C) = Charge (Q) / Potential difference (V)
C = Q/V
With the above formula, we can obtain the capacitance of the parallel plate capacitor as follow:
Potential difference (V) = 100 V.
Charge (Q) = 3.0×10¯⁹ C.
Capacitance (C) =..?
C = Q/V
C = 3.0×10¯⁹ / 100
C = 3.0×10¯¹¹ F.
Therefore, the capacitance of the parallel plate capacitor is 3.0×10¯¹¹ F.
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
You can use the impulse momentum theorem and just subtract the two momenta.
P1 - P2 = (16-1.2)(11.5e4)=1702000Ns
If you first worked out the force and integrated it over time the result is the same
Answer:
To find the mass using density and volume we just multiply them against each other which causes ml to cancel and just leaves us with grams which represents how much the item weights.



Therefore, our final answer is that our pencil weight 3.5 grams
<u><em>Hope this helps! Let me know if you have any questions</em></u>
Most likely the atoms of that elements are compress. :)