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Hitman42 [59]
1 year ago
9

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

quires 200 W of mechanical power. If we assume an efficiency similar to humans (say, 25%), a reasonable assumption, then the metabolic power of the swan is significantly higher than this. The swan does not stop to eat during a long day of flying; it get the energy it needs from fat stores. Assuming an efficiency similar to humans, after 12 hours of flight, a. How far has the swan traveled? b. What fraction of its body mass does it lose? Assume that I g of fat contains 37 kJ of energy.
Physics
1 answer:
Kisachek [45]1 year ago
4 0

Answer:

total distance moved by the swan is

d = 864000 m

Fraction of total mass lost by it

fraction = 0.0934

Explanation:

As we know that swan cruises at 20 m/s speed

Part a)

Distance moved by the swan is product of speed and time

so we have

d = v t

d = 20 \times (12 \times 3600)

d = 864000 m

Part b)

As we know that power consumed by the swan is 200 W

so total energy consumed in 12 hours is given as

E = P \times t

E = 200(12 \times 3600)

E = 8.64 \times 10^6 J

Since we know that output power is 25% of total energy used so total energy consumed is given as

E_{net} = 3.456 \times 10^7 J

now we know that 1 g of fat is lost for 37 kJ energy

so total mass lost in this

m = \frac{3.456\times 10^7}{37 \times 10^3}

m = 934 g

so fraction of mass lost is given as

fraction = \frac{0.934}{10}

fraction = 0.0934

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A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Bro
aleksley [76]

Answer:

Potential difference though which the electron was accelerated is 2.67\times 10^{-6}\ uV\  .

Explanation:

Given :

De Broglie wavelength , \lambda=750\ nm.

Plank's constant , h=6.626\times 10^{-34}\ J.s \ .

Charge of electron , e=-1.6\times 10^{-19}\ C.

Mass of electron , m=9.11\times 10^{-31}\ kg.m=9.11\times 10^{-31}\ kg.

We know , according to de broglie equation :

\lambda=\dfrac{h}{mv}\\\\ v=\dfrac{h}{m\lambda}\\\\v=\dfrac{6.626\times 10^{-34}\ J.s \ }{9.11\times 10^{-31}\ kg\times 750\times 10^{-9}\ m }= 969.78\ m/s .

Now , we know potential energy applied on electron will be equal to its kinetic energy .

Therefore ,

qV=\dfrac{mv^2}{2}\\\\ V=\dfrac{mv^2}{2q}

Putting all values in above equation we get ,

V=2.67\times 10^{-6}\ uV .

Hence , this is the required solution.

5 0
2 years ago
Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.
kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

q_{1} +q_{2} =-q_{3} -----eqution 1

where,

q_{1} is the heat absorbed by the solid at 0⁰C

q_{2} is the heat absorbed by the liquid at 0⁰C

q_{3} the heat lost by the warmer water sample

Important equations to be used in solving this problem

q=m *c*\delta {T}, where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

\delta {T} is change in temperature

Again,

q=n*\delta {_f_u_s} -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ

This means that equation (1) becomes

79.13 KJ + q_{2} = -q_{3}

<u>Step 4:</u> calculate the final temperature of the water

79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}

Substitute in the values; we will have,

79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})

79.13 kJ + 990.66J* (T_{f}-218}) = -1463J*(T_{f}-100})

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* (T_{f}-218}) = -1.463KJ*(T_{f}-100})

79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3

collect like terms,

2.45366T_{f} = 283.133

∴T_{f} = = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0
2 years ago
Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of
adoni [48]

Answer:

3311N

Explanation:

r = radius = 600m

V = speed = 150m/s

Mass = weight = 70kg

The weight of pilot when calculated due to circular motion

W = tv

Fv = mv²/r

Fv = 70x150²/600

Fv = 79x22500/600

= 15750000/600

= 2625N

Real Weight of the pilot = m x g

= 70 x 9.8

= 686N

The apparent Weight is calculated by

Mv²/r + mg

= 2625N + 686N

= 3311 N

Therefore the apparent Weight is 3311N

6 0
1 year ago
A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average
frez [133]
 (u) = 20 m/s 
(v) = 0 m/s 
<span> (t) = 4 s 
</span>
<span>0 = 20 + a(4) 

</span><span>4 x a = -20 
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>
8 0
2 years ago
Read 2 more answers
An air-filled capacitor is formed from two long conducting cylindrical shells that are coaxial and have radii of 30 mm and 80 mm
Licemer1 [7]

Answer:

24

Explanation:

4 0
1 year ago
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