A 2.0-kg projectile moves from its initial position to a point that is displaced 20 m horizontally and 15 m above its initial po

Answer:

Consequently, we find that:

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

both wires share the same length

both wires carry an identical current

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

We also know that where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

Thus, we conclude that

Now we are aware that the drift velocity of an electron in a wire can be described by:

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

So we observe that:

Thus, the most fitting answer is:

a. vA = vB/4

The full sentence states:
In a third class lever, the distance between the effort and the fulcrum is LESS than the distance between the load/resistance and the fulcrum.
In a third class lever, the fulcrum is positioned on one end of the effort, while the load/resistance is on the opposite side, placing the effort somewhere in between. Consequently, the distance from the effort to the fulcrum is less than that from the load to the fulcrum.

20.7 volts. The mass of an electron is 9.1 x 10⁻³¹ kg, and its wavelength is 0.27 x 10⁻⁹ m. The velocity of the electron can be determined using de Broglie's equation λ mv = h. Substituting the known values, we arrive at v = 2.7 x 10⁶ m/s. The potential difference through which the electron accelerates is noted, with the charge on an electron being 1.6 x 10⁻¹⁹ C. According to the conservation of energy, (0.5) mv² = q ΔV leads to ΔV = 20.7 volts.

B. red................

Answer:

10000 V

0.00225988700565 m²

Explanation:

E = Electric field =

d = Distance = 2.5 mm

Q = Charge = 80 nC

= Permittivity of free space =

The potential difference is calculated as

The potential difference across the plates amounts to 10000 V

Area is determined by

The area of each plate measures 0.00225988700565 m²

Capacitance is determined by

The capacitance is