The Correct Answer is <u>D.Infrared/</u> <em>INFARED has a lower frequency than visible light/</em>
Answer:
The electron tends to go to the region of 4. higher electric potential.
Explanation:
When a charged particle is immersed in an electric field, it experiences a force given by

where
q is the charge of the particle
E is the electric field
The direction of the force depends on the sign of the charge. In particular:
- The force and the electric field have the same direction if the charge is positive
- The force and the electric field have opposite directions if the charge is negative
Therefore, an electron (negative charge) moves in the direction opposite to the electric field lines.
However, electric field lines go from points at higher potential to points at lower potential: so, electrons move from regions at lower potential to regions of higher potential.
Therefore, the correct answer is
The electron tends to go to the region of 4. higher electric potential.
Answer:
PART A)
L = 97.4 dB
PART B)

PART C)
L = 67.9 dB
Explanation:
PART A)
level of sound is given as

now we have


PART B)
Since source is a spherical source
so here the intensity of sound is inversely depends on the square of the distance from the source



PART C)
level of sound is given as

now we have


The main parts of metabolism includes the two components which are: <em>Catholicism</em>, and <em>Anabolism</em>.
Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ