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AlladinOne [14]
3 years ago
5

Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, w

hich is 300 K. (Use 300 K as the temperature of the hot reservoir of the engine). The heat capacity of the brick is C
Physics
1 answer:
goblinko [34]3 years ago
5 0

Answer

Work done is 57.9KJ

Explanation

First solve the problem according to work done due to variation in temperature

So W= intergral Cu( 1-Tu/T). at Tu and T

So Given that

C = Heat capacity of the Brick

TEPc= Cold Temperature

TEPh = Hot Temperature

W = C ( TEPh-TEP) - TEPhCln ( TEPh/TEPc)

So

W= (1)-(300-150)-300 (1) ln 2

W= -57.9KJ

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Sam is walking at 2m/s down the hallway when he realized he will be late to class. He speeds up to 4m/s and makes it to English
Salsk061 [2.6K]

Answer:

.5 m/s²

Explanation:

a= (vf-vi)/t

a= (4-2)/4

a=2/4

a= .5 m/s²

8 0
3 years ago
You and some friends decide to take a canoe trip down the Wood River. The stretch of river you are traveling flows at a roughly
Anna35 [415]

Answer:

t= 1.2 hours

Explanation:

Define first di distance between the points, so

\bar{x}_{canoe}=2*(2+3)=10

\bar{x}_{water}=2*2=4

The distance is

d= \bar{x}_{canoe}- \bar{x}_{water}

d= 10-4 = 6miles

t = \frac{x}{v} = \frac{6}{2+3}

t= 1.2 hours

5 0
3 years ago
A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a
il63 [147K]

Answer:

0.0000109261200583 s

0.0109261200583

Explanation:

d_2 = Distance from right ear = 3 m

s = Distance between ears = 15 cm

v = Speed of sound in air = 343 m/s

Distance between the left ear and the bird

d_1=\sqrt{s^2+d_2^2}\\\Rightarrow d_1=\sqrt{0.15^2+3^2}\\\Rightarrow d_1=3.00374765918\ m=3.004\ m

Time

t=\dfrac{Distance}{Speed}

Time difference would be

\Delta T=\dfrac{d_1}{v}-\dfrac{d_2}{v}\\\Rightarrow \Delta T=\dfrac{3.00374765918}{343}-\dfrac{3}{343}\\\Rightarrow \Delta T=0.0000109261200583\ s

The time difference is 0.0000109261200583 s

Time period is given by

T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1000}\\\Rightarrow T=10^{-3}\ s

The ratio is

\dfrac{\Delta T}{T}=\dfrac{0.0000109261200583}{10^{-3}}\\\Rightarrow \dfrac{\Delta T}{T}=0.0109261200583

The ratio is 0.0109261200583

7 0
3 years ago
A 2500-kg car falls over a cliff and converts 1000,000J of PE into KE.
MArishka [77]

B is correct makes more sense

7 0
3 years ago
It's nighttime, and you ve dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90 m above the ed
yanalaym [24]

Answer:

The distance of the goggle from the edge is 5.30 m

Explanation:

Given:

The depth of pool (d) = 3.2 m

let 'i' be the angle of incidence

thus,

i = tan^{-1}(\frac{2.2}{0.90})

i = 67.75°

Now, Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

where,

r is the angle of refraction

n₁ is the refractive index of medium 1 = 1 for air

n₂ is the refractive index of medium 1 = 1.33 for water

now,

1 × sin 67.75° = 1.33 × sin(r)

or

r = 44.09°

Now,  

the distance of googles = 2.2 + d×tan(r)  = 2.2 + (3.2 × tan(44.09°) = 5.30 m

Hence, <u>the distance of the goggle from the edge is 5.30 m</u>

5 0
3 years ago
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