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Softa [21]
3 years ago
10

Calculate ideal work (in J) when a single stream of 1 mole of air is heated and expanded from 25 C and 1 bar to 100 C and 0.5 ba

r. The temperature of the surrounding is 37 C.
Note: Assume air to behave as an ideal gas under these conditions and ideal gas heat capacity is dependent on temperature]
Physics
1 answer:
NISA [10]3 years ago
7 0

Answer:

-1786.5J

Explanation:

Temperature 1=T1=25°c

Temperature 2=T2=200°c

Pressure P1=1bar

Pressure P2=0.5bars

T=37°c+273=310k

Note number if moles=1

Recall work done =2.3026RTlogp2/P1

2.3026*8.314*310log(0.5/1)

-1786.5J

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A submarine is 58.8 m from a whale. The sub sends out a sonar ping to locate the whale. The speed of sound underwater is 1520 m/
MissTica

Answer:

0.08

Explanation:

this problem assume that both of whale and submarine are in rest position or in constant linier motion in same direction and same speed.

The sound will travel from Submarine to the whale and back again to submarine. so the time will be like this

t = 2d/v

t = 2*58.8/1520

t = 117.6/1520

t = 0.077368 s

t ≈ 0.08 s (less then 1 s)

6 0
4 years ago
What is the most common consumed Halloween candy in the us after chocolate
iogann1982 [59]
Sour patch candy in the US after choclate
8 0
3 years ago
If your front lawn is 21.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1350 new snowflakes every minu
ziro4ka [17]

Answer:

snow is 64.638 kg / hr

Explanation:

Given data

wide w = 21 feet

long L = 20 ft

area A = 1350 square foot

mass of snow m  = 1.90 mg

to find out

snow in kilograms / hour

solution

we will find snow in kg

so we apply formula that is

snow kg / hour  = w × L ×A ×  m × 60/10^6

put all value we get  snow

snow =  21 × 20 × 1350 ×  1.90 × 60/10^6

snow =  420 × 1350 ×  1.90 × 60/10^6

snow =  1077300 × 60/10^6

snow =  64.638

hence snow is 64.638 kg / hr

7 0
3 years ago
while racing on a flat track, a car rounds a curve of 28m radius and instantaneously experiences a centripetal acceleration of 1
liubo4ka [24]
When a body strictly moves on a curve, it's velocity at a point is tangential to the curve at that point.

Centripetal acceleration is the acceleration that a body experiences by the virtue of change in it's tangential velocity. It is directed towards the centre and mathematically is v^2/R where v is the speed at the instant.

So, 18 = v^2/R
v^2 = 504
v = 6√14
4 0
3 years ago
A light spring stretches 0.13 m when a 0.35 kg mass is hung from it. The mass is pulled down from this equilibrium position an a
Alenkasestr [34]

Answer:

v = 1.30 m/s

Explanation:

given,

mass hung = 0.35 Kg

spring stretched when load is hanged  (x)= 0.13 m

now,

weight of the mass attached = Kx

             m g = k x

             0.35 x 9.8 = k x 0.13

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now, using conservation of energy

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 v = \sqrt{1.6958}

    v = 1.30 m/s

6 0
3 years ago
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