What do we call the movement of electric charge? A. Power B. Voltage C. Resistance D. Current

In an experiment to measure the acceleration due to gravity, g two values, 9.96 m/s2 and 9.72 m/s2 , are determined. Find (1) th

alexira [117]

Answer:

(1) Percent Difference = 2.47%

(2) Percent Error (9.96 m/s²) = 1.63 %

Percent Error (9.72 m/s²) = 0.82 %

(3) Percent Error (Mean) = 0.41 %

Explanation:

(1)

Percent Difference = [(9.96 m/s² - 9.72 m/s²)/(9.72 m/s²)]*100 %

Percent Difference = 2.47%

(2)

Percent Error = (|Measured Value - Original Value|/Original Value)*100%

Therefore,

Percent Error (9.96 m/s²) = (|9.96 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

Percent Error (9.96 m/s²) = 1.63 %

Now,

Percent Error (9.72 m/s²) = (|9.72 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

Percent Error (9.72 m/s²) = 0.82 %

(3)

First we need to find the mean of values:

Mean = (9.96 m/s² + 9.72 m/s²)/2

Mean = 9.84 m/s²

Therefore,

Percent Error (Mean) = (|9.84 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

Percent Error (Mean) = 0.41 %

8 0

3 years ago

A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in

Gekata [30.6K]

A.

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

Substituting the values of the variables into the equation, we have

Q = mcΔT

Q = ρVcΔT

Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C

Q = 482630400 J

So, the energy of the hot water is 482630400 J

B.

The elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

Substituting the values of the variables into the equation, we have

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.

Learn more about heat energy here:

brainly.com/question/11961649

5 0

3 years ago

Which of the following changes will always increase the efficiency of a thermodynamic engine? Choose all correct statements.

Ilya [14]

Answer:B,C,D

Explanation:

Thermodynamic efficiency is given by

$\eta =1-\frac{T_C}{T-H}$