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Pavel [41]
3 years ago
13

A 10 kg migratory swan cruises at 20m/s. A calculation that takes into ac-count the necessary forces shows that this motion requ

ires 200W of mechanical power. If we assume an efficiency similar to humans (say, 25%), a reasonableassumption, then the metabolic power of the swan is significantly higher thanthis. The swan does not stop to eat during a long day of flying; it get theenergy it needs from fat stores. Assuming an efficiency similar to humans, after12 hours of flight.
Required:
a. How far has the swan traveled?
b. How much metabolic energy has it used?
c. What fraction of its body mass does it lose?
Physics
1 answer:
IgorC [24]3 years ago
3 0

Answer:

Part A:

Distance=864000 m=864 km

Part B:

Energy Used=ΔE=8638000 Joules

Part C:

\frac{\triangle m}{m}=0.004998=0.49985\%

Explanation:

Given Data:

v=20m/s

Time =t=12 hours

In Secs:

Time=12*60*60=43200 secs

Solution:

Part A:

Distance = Speed**Time

Distance=v*t

Distance= 20*43200

Distance=864000 m=864 km

Part B:

Energy Used=ΔE= Energy Required-Kinetic Energy of swans

Energy Required to move= Power Required*time

Energy Required to move=200*43200=8640000 Joules

Kinetic Energy=\frac{1}{2}mv^2

K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules

Energy Used=ΔE=8640000 -2000

Energy Used=ΔE=8638000 Joules

Part C:

Fraction of Mass used=Δm/m

For This first calculate fraction of energy used:

Fraction of energy=ΔE/Energy required to move

ΔE is calculated in part B

Fraction of energy=8638000/8640000

Fraction of energy=0.99977

Kinetic Energy=\frac{1}{2}mv^2

Now, the relation between energies ratio and masses is:

\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2

\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977

\frac{\triangle m}{m}=0.004998=0.49985\%

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5 0
2 years ago
In a certain process, the energy change of the system is 250 \rm kJ. The process involves 480 \rm kJ of work done by the system.
alisha [4.7K]

Answer:

\Delta Q=-230kJ

Explanation:

Using the first law of thermodynamics:

\Delta U=\Delta Q-W

Where \Delta U is the change in the internal energy of the system, in this case  \Delta U=250kJ, \Delta Q is the heat tranferred, and W is the work,  W=-480kJ with a negative sign since the work is done by the system.

From the previous equation we solve for heat, because it is the unknown variable in this problem

\Delta Q=\Delta U +W

And replacing the known values:

\Delta Q=250kJ +(-480kJ)

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\Delta Q=-230kJ

The negative sign shows us that the heat is tranferred from the system into the surroundings.

3 0
3 years ago
How fast, in rpm, would a 5.6 kg, 25-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s
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Answer:

71 rpm

Explanation:

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Angular momentum (L) = 0.26

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Radius, r = (d/2) = 0.125m

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Moment of inertia (I) = 2mr² / 5

I = (2 * 5.6 * 0.125^2) / 5

= 0.175

= 0.175 / 5

= 0.035 kgm²

Angular speed (w) ;

w = L / I

w = 0.26 / 0.035

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w = 70.941724

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2 years ago
_____________ is the study of movement in athletes. A. Sports biomechanics B. Posture C. Dynamics D. Anatomy
liubo4ka [24]

Answer:

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3 0
2 years ago
What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

4 0
3 years ago
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