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Temka [501]
2 years ago
5

Consider a bird that flies at an average speed of 10.7 m/sm/s and releases energy from its body fat reserves at an average rate

of 3.70 WW (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g4.00g of fat to fly over a distance dbdbd_b without stopping for feeding. How far will the bird fly before feeding again
Physics
2 answers:
Katarina [22]2 years ago
7 0

Answer:

Part A: 455 km

Part B: 8.95 g

Part C: 3.11 g

Explanation:

A: Calculate distance by converting grams of fat to Joules, then using the equations t=W/P and d=v*t.

B: Calculate grams of carbohydrates by the same method.

C: Calculate grams of fat by the same method.

Wittaler [7]2 years ago
4 0

Answer:

455165.278 m

Explanation:

P = Power = 3.7 W

v = Velocity = 10.7 m/s

Amount of fat = 4 g

1 gram of fat provides about 9.40 (food) Calories

Energy given by 4 g of fat

E=4\times 9.4\times 4186\\\Rightarrow E=157393.6\ J

Time required to burn the fat

t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{157393.6}{3.7}\\\Rightarrow t=42538.811\ s

Distance traveled by the bird

s=vt\\\Rightarrow s=10.7\times 42538.811\\\Rightarrow s=455165.2777\ m

The bird will fly 455165.278 m

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Which of the following statements accurately describes the atmospheric patterns that influence local weather?
timurjin [86]

Answer: A

Explanation:

Well the high and lows effect the humidity the more humidity the more hot it is so the high brings higher temperatures.

4 0
1 year ago
Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the gener
Andrej [43]

Answer:

( a ) The specific volume by ideal gas equation = 0.02632 \frac{m^{3} }{kg}

% Error =  20.75 %

(b) The value of specific volume From the generalized compressibility chart = 0.0142 \frac{m^{3} }{kg}

% Error =  - 34.85 %

Explanation:

Pressure = 1 M pa

Temperature = 50 °c = 323 K

Gas constant ( R ) for refrigerant = 81.49 \frac{J}{kg k}

(a). From ideal gas equation P V = m R T ---------- (1)

⇒ \frac{V}{m} = \frac{R T}{P}

⇒ Here \frac{V}{m} = Specific volume = v

⇒ v =  \frac{R T}{P}

Put all the values in the above formula we get

⇒ v = \frac{323}{10^{6} } ×81.49

⇒ v = 0.02632 \frac{m^{3} }{kg}

This is the specific volume by ideal gas equation.

Actual value = 0.021796 \frac{m^{3} }{kg}

Error =  0.02632 - 0.021796 =   0.004524 \frac{m^{3} }{kg}

% Error =  \frac{0.004524}{0.021796} × 100

% Error =  20.75 %

(b). From the generalized compressibility chart the value of specific volume

 \frac{V}{m} = v = 0.0142 \frac{m^{3} }{kg}

The actual value = 0.021796 \frac{m^{3} }{kg}

Error = 0.0142 - 0.021796 =  \frac{m^{3} }{kg}

% Error = \frac{- 0.0076}{0.021796} × 100

% Error =  - 34.85 %

3 0
2 years ago
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
saw5 [17]

Answer:

The electric field strength is 4.5\times 10^{4} N/C

Solution:

As per the question:

Area of the electrode, A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}

Charge, q = 50 nC = 50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}

Now, the electric field strength of the electrode is:

\vec{E} = \frac{\sigma}{2\epsilon_{o}}

where

\epsilon_{o} = 8.85\times 10^{- 12} F/m

\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}

\vec{E} = 4.5\times 10^{4} N/C

7 0
2 years ago
In an experiment to measure the wavelength of light using a double slit, it is found that the fringes are too close together to
Mandarinka [93]

Answer:

halve the slit separation

Explanation:

As we know that

In YDS experiment, the equation of fringe width is as follows

\beta = \frac{\lambda D}{d}

where,

D denotes the separation in the middle of screen and slits

d denotes the distance in the middle of two slits

And to increase the Δx we have to decrease the d i.e, the distance between the two slits

Hence, the first option is correct

4 0
2 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
LuckyWell [14K]

For a circular aperture, the first minima (n=1) as an angular separation from the peak of the central maxima given by

Sinθ = 1.22λ / d

Where,

d is the aperture or pupil diameter

d = 4.69 mm = 4.69 × 10^-3m

λ is the wavelength

λ = 545 nm = 545 × 10^-9 m

Then,

Sinθ = 1.22λ / d

Sinθ = 1.22 × 545 × 10^-9 / 4.69 × 10^-3

Sinθ = 1.418 × 10^-4 rad

Then, the head light sources have the same angular separation θ from the eye as the image have inside the eye.

For the headlight

Sinθ ≈ light separation / distantce for the eye

Light separation is give as x = 0.659 m

And let the distance of the eye be D

Then,

Sinθ = x / D

Make D subject of formula

D = x / Sinθ

D = 0.695 / 1.418 × 10^-4

D = 4902.316m

To km, 1km = 1000m

D ≈ 4.9 km

4 0
2 years ago
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