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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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4 years ago
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2, separated by a distance of 1.70 mm. A 2
Aleks04 [339]

Answer:

(a) 1.47 x 10⁴ V/m

(b) 1.28 x 10⁻⁷C/m²

(c) 3.9 x 10⁻¹²F

(d) 9.75 x 10⁻¹¹C

Explanation:

(a) For a parallel plate capacitor, the electric field E between the plates is given by;

E = V / d               -----------(i)

Where;

V = potential difference applied to the plates

d = distance between these plates

From the question;

V = 25.0V

d = 1.70mm = 0.0017m

Substitute these values into equation (i) as follows;

E = 25.0 / 0.0017

E = 1.47 x 10⁴ V/m

(c) The capacitance of the capacitor is given by

C = Aε₀ / d

Where

C = capacitance

A = Area of the plates = 7.60cm² = 0.00076m²

ε₀ = permittivity of free space =  8.85 x 10⁻¹²F/m

d = 1.70mm = 0.0017m

C = 0.00076 x  8.85 x 10⁻¹² / 0.0017

C = 3.9 x 10⁻¹²F

(d) The charge, Q, on each plate can be found as follows;

Q = C V

Q =  3.9 x 10⁻¹² x 25.0

Q = 9.75 x 10⁻¹¹C

Now since we have found other quantities, it is way easier to find the surface charge density.

(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e

σ = Q / A

σ = 9.75 x 10⁻¹¹ /  0.00076

σ = 1.28 x 10⁻⁷C/m²

6 0
4 years ago
Show your work.. to
stiks02 [169]
SOLUTION:

Question 4 -

- 2x + 5 = - 13 + x

- 2x - x = - 13 - 5

- 3x = - 18

x = - 18 / - 3

x = 18 / 3

x = 6

Question 6 -

2x + 7 = - 2 - x

2x + x = - 2 - 7

3x = - 9

x = - 9 / 3

x = - 3

Question 8 -

2x + 12 = - 6 - x

2x + x = - 6 - 12

3x = - 18

x = - 18 / 3

x = - 6

Hope this helps! :)
Have a lovely day! <3
5 0
4 years ago
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3 years ago
A squirrel pulls a 8.7 kg bag of acorns along the ground with a constant force of 80 N. If the force of friction opposing this s
Annette [7]

Answer:

a=5.51\ m/s^2

Explanation:

Given that,

The mass of a bag of acorns, m = 8.5 kg

The force acting on the bag = 80 N

The force of friction = 32 N

We need to find the acceleration of the bag of acorns.

Net force acting on it is = 80 N - 32 N

= 48 N

Let a be the acceleration of the bag of acorns. So,

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{48}{8.7}\\\\a=5.51\ m/s^2

So, the acceleration of the bag of acorns is 5.51\ m/s^2.

5 0
3 years ago
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