What should you do immediately if a boat motor catches fire?
1 answer:
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The energy stored in the membrane is
Explanation:
The capacitance of a parallel-plate capacitor is given by
where
k is the dielectric constant of the material
is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
For the membrane in this problem, we have
k = 4.6
Substituting, we find its capacitance:
Now we can find the energy stored: for a capacitor, it is given by
where
is the capacitance
is the potential difference
Substituting,
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Answer:
a. It starts at point B.
vp = 2.53*10⁴ m/s
a. it starts at point A.
ve= 1.08*10⁶ m/s
Explanation:
a) As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.
Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:
ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)
⇒
where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.
Replacing by these values, and solving for v, we have:
⇒ vp = 2.53*10⁴ m/s
b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:
First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.
Second, as its charge is (-e) the change in electric potential energy had been negative also:
ΔUe = -e*ΔV = -e* (VB-VA)
In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:
where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.
Replacing by these values, and solving for v, we have:
⇒ ve = 1.08*10⁶ m/s
Answer:
The correct option is;
Man doing most work
Box gaining the most energy
D Y Q
Explanation:
The given parameters of the question are;
The distance the box P is pushed by the Man X = 0
The force the Man X applies to the box P = x N
The distance the box Q is lifted by the Man Y = h > 0 meters
The minimum force the Man Y applies to the box Q = W, the weight of the box
Work done = Force × Distance
Energy gained = Potential energy + Kinetic Energy = (Mass × Gravity × Height = Weight × Height = W × h) + 1/2 × Mass × Velocity²
The final velocity of either box = 0 m/s (The boxes are at rest on the ground or the shelf)
Therefore, Kinetic energy = 0 J
The work done by Man X = 0 × x = 0 J
The energy gained by the box P = W × 0 = 0 J
The work done by Man Y = W × h = W·h J
The energy gained by the box P = W × h = W·h J
We have, the work done by the man Y = W·h J is more than the work done by the man X = 0 J
The energy gained by the box P = W·h J is more than the energy gained by the box Q = 0 J
Therefore, the correct option is D, Man doing the most work is Y, box gaining the most energy is Q.