The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.
Location ' x ' is √(2² + 3²) = √13 m from the charge.
Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.
The magnitude of the E-field is the same at both locations.
The direction is also the same at both locations ... it points toward the origin.
The pressure at the depth 11 km below sea level can be
calculated using
P=ρgh
P is pressure, ρ is the density of the fluid; g is the
gravitational constant, h is the height from the surface, or depth that the
object is submerged.
P = ( 1000 kg/ m3) ( 9.81 m.s2)( 11 000m) + 1 atm
P = 107,910,000 pa ( 1 atm/ 101 325 Pa) + 1 atm = 1066 atm
Answer:
= 1.75 × 10⁻⁴ m/s
Explanation:
Given:
Density of copper, ρ = 8.93 g/cm³
mass, M = 63.5 g/mol
Radius of wire = 0.625 mm
Current, I = 3A
Area of the wire, 
= 
Now,
The current density, J is given as
= 2444619.925 A/mm²
now, the electron density, 
where,
=Avogadro's Number
Now,
the drift velocity,
where,
e = charge on electron = 1.6 × 10⁻¹⁹ C
thus,
= 1.75 × 10⁻⁴ m/s
Answer:
his results in the final angle after the collision of 37.2 degrees basically what we did there is turn the vector into a right triangle. We use sohcahtoa to solve for the angle. Being.
Explanation:
Answer b protons and electrons
