The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
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Answer:
2.24 seconds
Explanation:
xf = xo + vo t + 1/2 at^2
45 = 0 + 15 t + 1/2 (4.5) t^2
2.25 t^2 + 15t - 45 = 0 Quadratic formula shows t = 2.24 seconds
During the "U" part of the turn, the car would follow an approximately circular path, and if it's moving at a constant speed, it would have to accelerate toward the center of the circle in order to change its direction.
The magnetic field at the center of the arc is 4 × 10^(-4) T.
To find the answer, we need to know about the magnetic field due to a circular arc.
<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>
- According to Biot savert's law, magnetic field at the center of a circular arc is
- B=(μ₀ I/4π)× (arc/radius²)
- As arc is given as angle × radius, so
B=( μ₀I/4π)×(angle/radius)
<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>
B=(μ₀ I/4π)× (0.9/0.006)
= (10^(-7)× 26.9)× (0.9/0.006)
= 4 × 10^(-4) T
Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.
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A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).
A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:
- From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
- From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.
We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:
- The speed increases from 0 m/s to 1341 m/s.
- The time elpased is 2.0 min.
- 1 min = 60 s.
The acceleration of the ship during the first 2.0 minutes is:
A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).
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