Answer:
(D) energy from one place to another
Answer:
R = 0.992 Ω
Explanation:
En esta pregunta, dada la información que contiene, debemos calcular la resistencia de la varilla de grafito.
Matemáticamente,
Resistencia = (resistividad * longitud) / Área De la pregunta;
Resistividad = 3,5 * 10 ^ -5 Ωm
longitud = 170 cm = 1,7 m
Área = 60 mm ^ 2 = 60/1000000 = 6 * 10 ^ -5 m ^ 2
Conectando estos valores a la ecuación anterior, tenemos;
Resistencia = (3.5 * 10 ^ -5 * 1.7) / (6 * 10 ^ -5) =
(3.5 * 1.7) / 6 = 0.992 Ω
<span>One thousand grams of seawater has 35 grams of dissolved substances ... on the average. While the salinity of the Earth's oceans and seas varies, the average salinity of seawater rests at 3.5%. Consider one liter or sea or ocean water. One liter has 1,000 milliliters (mL) in it. To find 3.5% of 1,000, we would multiply with the decimal place adjusted for percentages: 1000 x .035 = 35. Therefore, for every 1,000 mL of seawater, we will find 35 grams of (mostly) sodium chloride, otherwise known as salt.</span>
Answer:
D. A negative externality.
Explanation:
Particulate pollution is a form of pollution that is responsible for the degradation of the environment.
Particulate matter is also referred to as particle pollution or atmospheric aerosol particles and it can be defined as a complex microscopic mixture of liquid droplets and solid particles that are suspended in air.
An externality is typically an unwarranted cost or benefit by a manufacturer or producer of goods and services that affects a third party.
In Economics, an externality could either be positive or negative depending on its effect on a third party.
A negative externality arises when the production or consumption of a finished product or service has negative impact (cost) on a third party.
In conclusion, air pollution is an example of a negative externality because it causes harm to a third party.
Answer:
(a) 333.77 J
(b) 237.85 J
(c) 4763.77 J
(d) 4667.85 J
Explanation:
Temperature of source, TH = 314 K
Temperature of A, Tc = 292 K
Temperature of B, Tc' = 298 K
heat taken out, Qc = 4430 J
Let the heat deposited outside is QH and QH' by A and B respectively.

Now

(a) Work done for A
W = QH - QC = 4763.77 - 4430 = 333.77 J
(b) Work done for B
W' = QH' - Qc = 4667.85 - 4430 = 237.85 J
(c) QH = 4763.77 J
(d) QH' = 4667.85 J