Cuales son las carreras de velocidad de medio fondo y fondo?

Part C Let’s start the analysis by looking at your “extreme usage” cases. Compare the two cases in detail—low usage period versu

Serga [27]

Answer:

Day 7 DataUsage notes (since last reading)day & datetimekWh readingkWh usedhours elapsedavg. kW usedb.Usage Extremes: Data CollectionFor this experiment, you’ll measure electrical usage during a time period when you expect to havevery light electrical usage (for instance, while you’re asleep at night or during the day when no oneis at home). Likewise you’ll measure electrical usage during a time period when you expect to have heavier than average electrical usage. This time period might be in the evening, when lights and other appliances are on. Both of these time periods should be at least 4 hours long, to increase the accuracy of your results. Record your results in the tables below for each situation. For each time period, you’ll need to takean initial and a final reading.Type your response here:Low Usage - Initial Readingday & datetimekWh readingLow Usage - Final ReadingEnergy Usage Notesday & datetimekWh readingkWh usedhours elapsedavg. kW usedHigh Usage - Initial Readingday & datetimekWh reading4

6 0

2 years ago

Capacitances of 10uF and 20uF are connected in parallel,

jarptica [38.1K]

Answer:

The equivalent capacitance will be $15\mu F$

Explanation:

We have given two capacitance $C_=10\mu F\ and\ C_2=20\mu F$

They are connected in parallel

So equivalent capacitance $C=C_1+C_2=10+20=30\mu F$

This equivalent capacitance is now connected in series with $30\mu F$

In series combination of capacitors the equivalent capacitance is given by $\frac{1}{C}=\frac{1}{30}+\frac{1}{30}$

$C=\frac{30}{2}=15\mu F$

So the equivalent capacitance will be $15\mu F$

5 0

2 years ago

A 75.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.79 de

jeyben [28]

Answer:

$c_{e1}$ = 0.331 J / g ° C

Explanation:

We have a calorimetry exercise where all the heat yielded by one of the components is absorbed by the other.

Heat ceded Qh = m1 ce1 ( $T_{h}$ - $T_{f}$ )

Heat absorbed Qc = m2 ce2 ( $T_{f}$ - T₀)

Body 1 is metal and body 2 is water . Where m are the masses of the two bodies, ce their specific heat and T the temperatures

Qh = Qc

m₁ $c_{e1}$ ( $T_{h}$ - $T_{f}$ ) = m₂ $c_{e2}$ ( $T_{f}$ - T₀)

we clear the specific heat of the metal

$c_{e1}$ = m₂ $c_{e2}$ ( $T_{f}$ - T₀) / (m₁ ( $T_{h}$ - $T_{f}$ ))

$c_{e1}$ = 50.00 4.184 (20.15 -10.79) / (75.00 (99.0-20.15))

$c_{e1}$ = 209.2 (9.36) / (75 78.85)

$c_{e1}$ = 1958.11 / 5913.75

$c_{e1}$ = 0.331 J / g ° C

5 0

3 years ago

Find the average energy E for (a) An n-state system, in which a given state can have energy 0,,2,...,n . (b) A harmonic oscillat

kicyunya [14]

Answer:

a♦1 E_average = n E₀ / 2 , b) E_average= infinity

Explanation:

The energy values form an arithmetic series, whose sum is

S = n (a₁ + aₙ) / 2 = n (2a₁ + (n-1) r)/ 2

Where n is the number of terms, a₁ is the first term, aₙ the last term and r is the difference between two consecutive numbers in the series

r = 2E₀ - 0 = 2E₀

Therefore the sum is

S = n (0 + n E₀) / 2

S = n² E₀ / 2

The average value is

E_average = S / n

E_average = n E₀ / 2

b) the case of harmonic oscillation

We have two possibilities.

- if we take a finite number and terms gives the same previous value

- If we take an infinite number of fears the series gives infinity and the average is also infinite

E_average= infinity

4 0

3 years ago

How does dance use energy to influence motion?

Goshia [24]

Solar system energy. electoral power engery motion force engery . wind energy. kinetic energy

8 0

2 years ago