Air expands as it warms. Therefore warm air is less dense than cool air. The warm air from the first floor apartments rises to the second floor. People on the second floor require less heating to keep their <span>apartments comfortable.
-Hope this helps </span>
If the solution is treated as an ideal solution, the extent of freezing
point depression depends only on the solute concentration that can be
estimated by a simple linear relationship with the cryoscopic constant:
ΔTF = KF · m · i
ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF
(solution).
KF, the cryoscopic constant, which is dependent on the properties of the
solvent, not the solute. Note: When conducting experiments, a higher KF
value makes it easier to observe larger drops in the freezing point.
For water, KF = 1.853 K·kg/mol.[1]
m is the molality (mol solute per kg of solvent)
i is the van 't Hoff factor (number of solute particles per mol, e.g. i =
2 for NaCl).
Answer: The copper wiring most likely caused the mini-refrigerator fire.
Explanation: A mini-refrigerator may have one or more evaporator fan motors. These motors have copper wiring. In some situation these copper wires gets overheated. It may result to the firing of the mini-refrigerator.
Therefore, the mini-refrigerator fire was most likely caused by copper wiring.
Answer:
A. The electron will begin to move along the axis, towards the centre and the instantaneous velocity because the force acting on it depends largely on acceleration and x until it reaches maximum velocity at centre.
B. Veloctiy (Vb) = 1.66m/s
Explanation:
Given the following data
x(a) = 0.3m
x(b) = 0
q = 1.6×10^-19
Q = 24nc
r = 0.15m
Required: the motion of the electron and the velocity (Vb)
1. At point A the electron will begin to move along the axis from point A to point B, the magnitude of the electric field will change while moving which depends on that and this will produce instantaneous force which will later change and the acceleration will change too while moving, the velocity would reach maximum value at point B
2. Potential energy and kinetic energy are given by
U(a) + K(a) = U(b) + K(b). . .1
Initial P.E and K.E are given as
U(a) = kQ/√x²(a) + a2
By substitution, we have
U(a) = 9×10^9 × (-1.9×10^-19)×24×10^-9/√(0.15)²+(0.3²)
U(a) = -1.03×10^-16
Final P.E and K.E are given as
U(b) = KQ/√x²(b) + a2
By substitution, we have
U(b) = 9×10^9×(-1.9×10^-19)×24×10^-9/√(0.15)²+(0)²
U(b) = -2.3×10^16
3. By substitution into equation 1 becomes
-1.03×10^-6 - 2.3×10^-16 + MV²(b)/2
V(b) = √2×1.27×10^-16/9.1×10^31
V(b) = 1.66×10^7m/s
