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dangina
2 months ago
15

Define a named tuple Player that describes an athlete on a sports team. Include the fields name, number, position, and team.

Engineering
1 answer:
choli [298]2 months ago
7 0

Respuesta:

Explicación:

analizaremos esta pregunta con el código relevante proporcionado

#código:

from collections import namedtuple

#creando una tupla nombrada 'Player' con los campos name, number, position y team

Player = namedtuple('Player',['name','number','position', 'team'])

cam = Player('Cam Newton','1','Quarterback','Carolina Panthers')

lebron = Player('Lebron James','23','Small forward','Los Angeles Lakers')

print(cam.name+'(#'+cam.number+')'+' es un '+cam.position+' para el '+cam.team+'.')

print(lebron.name+'(#'+lebron.number+')'+' es un '+lebron.position+' para el '+lebron.team+'.')

NB:

Lebron James (#23) es el alero de Los Angeles Lakers

Cam Newton(#1) es el mariscal de campo de los Carolina Panthers

saludos, espero que esto ayude

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Volume change percentage is 2.60%

Water level increase is 4.138 mm

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Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

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We aim to determine percentage change in volume and the rise in water level.

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We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

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E = -\frac{dp}{dV/V}................1

And -\frac{dV}{V} = \frac{d\rho}{\rho}............2

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The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

-\frac{dV}{V} = \frac{d\rho}{\rho}

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Therefore, the percentage change in volume will be

dV % = -\frac{dV}{V}  × 100

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dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = \frac{\pi }{4} *d^2*l(i)................3

Final volume v2 = \frac{\pi }{4} *d^2*l(f)................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 =  \frac{\pi }{4} *d^2*(l(f)-l(i))

dV = \frac{\pi }{4} *d^2*dl

Substituting all values yields

0.0130 = \frac{\pi }{4} *2^2*dl

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

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