Answer:
Follows are the solution to this question:
Explanation:
Calculating the area under the curve:
A = as
Calculating the kinematics equation:
Calculating the value of acceleration:
On a hot summer day, a student turns his fan on when he leaves his room in the morning. When he returns in the evening, will the
1 answer:
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Answer:
Shining lantern Magic beans
Explanation:
I will explain the code line by line.
The first statement contains a string type variable magicPowers and second statement has an int type variable experienceLevel which is initialized by value 9.
Nested if statement are used and lets see which of those statements execute.
The first if statement checks if the value of experienceLevel is greater than 10. This statement evaluates to false because the value of experienceLevel is set to 9. Then the program control moves to the next if statement.
The second if statement checks if the value of experienceLevel is greater than 8. This if condition evaluates to true because the value of experienceLevel is 9. So this if statement is executed which contains the statement magicPowers = magicPowers + "Shining lantern "; . This means magicPowers stores the string Shining lantern.
The third if statement checks if the value of experienceLevel is greater than 2. As the value of experienceLevel is 9 so this condition also evaluates to true which means the statement magicPowers = magicPowers + "Magic beans "; is executed. This statement means magicPowers stores the string Magic Beans
You can use a print statement to display the output on the screen as:
System.out.print(magicPowers);
Output:
Shining lantern Magic beans
The program along with the output is attached as a screenshot.
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