Answer:
the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Explanation:
Given the data in the question;
we make use of the following expression;
hall Voltage VH = IB / ned
where I = 2.25 A
B = 0.685 T
d = 0.107 mm = 0.107 × 10⁻³ m
e = 1.602×10⁻¹⁹ C
VH = 2.59 mV = 2.59 × 10⁻³ volt
n is the electron density
so from the form; VH = IB / ned
VHned = IB
n = IB / VHed
so we substitute
n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )
n = 1.54125 / 4.4396226 × 10⁻²⁶
n = 3.4716 × 10²⁵ m⁻³
Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Answer: 361° C
Explanation:
Given
Initial pressure of the gas, P1 = 294 kPa
Final pressure of the gas, P2 = 500 kPa
Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K
Final temperature of the gas, T2 = ?
Let us assume that the gas is an ideal gas, then we use the equation below to solve
T2/T1 = P2/P1
T2 = T1 * (P2/P1)
T2 = (100 + 273) * (500 / 294)
T2 = 373 * (500 / 294)
T2 = 373 * 1.7
T2 = 634 K
T2 = 634 K - 273 K = 361° C
The answer is C.
The Kinetic energy which was exerted and experience pulling the string of a bow is kept as a potential energy at the end of the arrow in contact with the string. Once release from aim at stationary position the potential energy is again transformed.
Answer:
3 electron hai bro of puch mujhe sab aata h
