What is a circuit that only has one loop??

Flapping flight is very energy intensive. A wind tunnel test

steposvetlana [31]

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight= $\frac{P}{m}$

Using the formula

Metabolic power for starting flight= $\frac{12}{0.089}$

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

4 0

3 years ago

HELP HELP HELP help pls

GalinKa [24]

Answer is A.) 40 m/km

7 0

2 years ago

Which of the following is true about Viscosity of liquids:

White raven [17]

Viscosity of liquids is essentially the 'thickness' of the liquid. For instance, honey and water have different viscosities. Honey has a higher one and therefore, liquids with high viscosity do not flow as well as liquids with low viscosity (water).

4 0

2 years ago

Read 2 more answers

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

4 0

3 years ago

A sample of an unknown substance has a mass of 0. 158 kg. If 2,510. 0 J of heat is required to heat the substance from 32. 0Â°C

guapka [62]

The specific heat of the unknown substance with a mass of 0.158kg is 0.5478 J/g°C

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

The specific heat capacity of a substance can be calculated using the following formula:

Q = m × c × ∆T

Where;

Q = quantity of heat absorbed (J)
c = specific heat capacity (4.18 J/g°C)
m = mass of substance
∆T = change in temperature (°C)

According to this question, 2,510.0 J of heat is required to heat the 0.158kg substance from 32.0°C to 61.0°C. The specific heat capacity can be calculated:

2510 = 158 × c × (61°C - 32°C)

2510 = 4582c

c = 2510 ÷ 4582

c = 0.5478 J/g°C

Therefore, the specific heat capacity of the unknown substance that has a mass of 0.158 kg is 0.5478 J/g°C.

Learn more about specific heat capacity at: brainly.com/question/2530523

4 0

2 years ago