Answer:
T_{f} = 90.07998 ° C
Explanation:
This is a calorimetry process where the heat given by the Te is absorbed by the air at room temperature (T₀ = 25ºC) with a specific heat of 1,009 J / kg ºC, we assume that the amount of Tea in the cup is V₀ = 100 ml. The bottle being thermally insulated does not intervene in the process
Qc = -Qb
M 
(T₁ -
) = m 
(T_{f}-T₀)
Where M is the mass of Tea that remains after taking out the cup, the density of Te is the density of water plus the solids dissolved in them, the approximate values are from 1020 to 1200 kg / m³, for this calculation we use 1100 kg / m³
ρ = m / V
V = 1000 -100 = 900 ml
V = 0.900 l (1 m3 / 1000 l) = 0.900 10⁻³ m³
V_air = 0.100 l = 0.1 10⁻³ m³
Tea Mass
M = ρ V_te
M = 1100 0.9 10⁻³
M = 0.990 kg
Air mass
m = ρ _air V_air
m = 1.225 0.1 10⁻³
m = 0.1225 10⁻³ kg
(m c_{e_air} + M c_{e_Te}) T_{f}. = M c_{e_Te} T1 - m c_{e_air} T₀
T_{f} = (M c_{e_Te} T₁ - m c_{e_air} T₀) / (m c_{e_air} + M c_{e_Te})
Let's calculate
T_{f} = (0.990 1100 90.08– 0.1225 10⁻³ 1.225 25) / (0.1225 10⁻³ 1.225 + 0.990 1100)
T_{f} = (98097.12 -3.75 10⁻³) / (0.15 10⁻³ +1089)
T_{f} = 98097.11 / 1089.0002
T_{f} = 90.07998 ° C
This temperature decrease is very small and cannot be measured
Answer:
A. Education
Explanation:
Having an education allows a person to learn certain set of skills and knowledge around the world. This may act as a "buffer" or a barrier against unemployment because<u> it gives the person a higher chance of getting employed.</u> Many people know that possessing a certain degree gives you an edge to get interviewed and if possible, get hired. Continuous learning (education) makes you a <u>well-rounded person</u> capable of doing just any kind of job available out there.
So, this explains the answer.
Answer:
A RCRA characteristic hazardous waste is a solid waste that exhibits at least one of four characteristics defined in 40 CFR Part 261 subpart C — ignitability (D001), corrosivity (D002), reactivity (D003), and toxicity (D004 - D043). combustible, or have a flash point less than 60 °C (140 °F).
Answer:
2.9*10^14 electrons
Explanation:
An Ampere is the flow of one Coulomb per second, so 35 μA = is 35*10^-6 C per second.
An electron has a charge of 1.6*10^-19 C.
35*10^-6 / 1.6*10^-19 = 2.9*10^14 electrons
So, with a current o 35 μA you have an aevrage of 2.9*10^14 electrons flowing past a fixed reference cross section perpendicular to the direction of flow.
Answer:
import java.util.Scanner;
public class FindMatchValue {
public static void main (String [] args) {
Scanner scnr = new Scanner(System.in);
final int NUM_VALS = 4;
int[] userValues = new int[NUM_VALS];
int i;
int matchValue;
int numMatches = -99; // Assign numMatches with 0 before your for loop
matchValue = scnr.nextInt();
for (i = 0; i < userValues.length; ++i) {
userValues[i] = scnr.nextInt();
}
/* Your solution goes here */
numMatches = 0;
for (i = 0; i < userValues.length; ++i) {
if(userValues[i] == matchValue) {
numMatches++;
}
}
System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);
}
}
