Answer:
A. Sedimentary
Explanation:
I took the test and got this correct. Hopefully this helps you!
We know that the average speed is simply the ratio of the
total distance travelled over the total duration of the trip.
total distance = 500 mi + 380 mi + 600 mi
total distance = 1,480 mi
total time = 10 h + 8 h + 15 h
total time = 33 h
So the average speed is therefore:
average speed = 1,480 mi / 33 h
<span>average speed = 44.85 mi / h</span>
The force on a charged particle in a magnetic field is given by
the speed of the charged particle = 10842 m/s.
Explanation:
F= q V B sinθ
F=force=3.5 x 10⁻²N
q= charge= 8.4 x 10⁻⁴ C
B= magnetic field= 6.7 x 10⁻³ T
θ=35⁰
Thus the velocity is given by V=
V=(3.5 x 10⁻²)/[(8.4 x 10⁻⁴)(6.7 x 10⁻³)(sin35)]
V=10842 m/s
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
Learn more about potential difference across parallel plate capacitor here:
brainly.com/question/12993474
Answer:
Explanation:
A grounded wire is sometimes strung along the tops of the towers to provide lightning protection.
In areas where the neutral is grounded or earthed, it is essential to endure that the neutral and the live or hot wires are not confused for each other.
When this happens, the fuses on the transformer will not operate unless the fault is very close to the transformer. The fuses in the consumer's intake box, will not operate.
