1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Alchen [17]
3 years ago
15

an engine has been design to work between source and the sink at temperature 177 degree Celsius and 27 degree Celsius respective

ly. if the energy input is 3600 joule. what is the work done by the engine
Physics
1 answer:
irina [24]3 years ago
4 0

<u>Given data</u>

Source temperature (T₁) = 177°C = 177+273 = 450 K

Sink temperature (T₂) = 27°C = 27+273 = 300 K

Energy input (Q₁) = 3600 J ,

Work done = ?

                We know that, efficiency (η) = Net work done ÷ Heat supplied

                                                           η =   W ÷ Q₁  

                                                           W = η × Q₁

               First determine the efficiency ( η ) = ?

                                Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)

                                                                        = 33.3% = 0.333

               Now, Work done is W = η × Q₁

                                                    = 0.33 × 3600

                                                 <em>  W = 1188 J</em>

<em>Work done by the engine is 1188 J</em>

You might be interested in
What happens to its kinetic energy when its reaches the maximum height ?​
drek231 [11]

when it reaches the maximum height, all the energy has now been converted into potential energy.when a ball is thrown straight upto into the air,all its initial kinetic energy converted into gravitational potential energy when it reaches its maximum height

7 0
3 years ago
A fairground ride consists of a large vertical drum that spins so
expeople1 [14]

Answer:

v = 3.84 m/s

Explanation:

In order for the riders to stay pinned against the inside of the drum the frictional force on them must be equal to the centripetal force:

Centripetal\ Force = Frictional\ Force\\\\\frac{mv^2}{r} = \mu R = \mu W\\\\\frac{mv^2}{r} = \mu mg\\\\\frac{v^2}{r} = \mu g\\\\v = \sqrt{\mu gr}

where,

v = minimum speed = ?

g = acceleration due to gravity = 9.81 m/s²

r = radius = 10 m

μ = coefficient of friction = 0.15

Therefore,

v=\sqrt{(0.15)(9.81\ m/s^2)(10\ m)}

<u>v = 3.84 m/s</u>

6 0
3 years ago
A 1300-kg car initially has a velocity of 22.2 m/s due south. It brakes to a stop over a 180 m distance.
Vaselesa [24]
The acceleration of the object which moves from an initial step to a full halt given the distance traveled can be calculated through the equation,
                                     d = v² / 2a
where d is distance, v is the velocity, and a is acceleration
Substituting the known values,
                                     180 = (22.2 m/s)² / 2(a)
The value of a is equal to 1.369 m/s²
The force needed for the object to be stopped is equal to the product of the mass and the acceleration.
                                      F = (1300 kg)(1.369 m/s²) 
                                            F = 1779.7 N
4 0
3 years ago
How to determine whether the object is magnetic.
o-na [289]

Answer:

By holding another magnet close to it. If the object is attracted to the magnet, then it too is magnetic.

7 0
3 years ago
Read 2 more answers
The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha
belka [17]

Given that,

Energy H=2.7\times10^{31}\ W

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

H=Ae\sigma T^4

A=\dfrac{H}{e\sigma T^4}

4\pi R^2=\dfrac{H}{e\sigma T^4}

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}

R=5.0\times10^{10}\ m

(b). Given that,

Radiates energy H=2.1\times10^{23}\ W

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}

R=5.42\times10^{6}\ m

Hence, (a). The radius of the star is 5.0\times10^{10}\ m

(b). The radius of the star is 5.42\times10^{6}\ m

8 0
3 years ago
Other questions:
  • A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a
    8·1 answer
  • The captain told the passengers that the plane is flying at 450 miles per hour this information describes the plane
    6·2 answers
  • A mixed cost contains Select one:
    8·1 answer
  • A large truck collides head-on with a small compact car. During the collision: (A) the truck exerts a greater amount of force on
    9·1 answer
  • Please, please, please help me!!!!!!
    9·1 answer
  • The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t
    14·1 answer
  • A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to
    7·1 answer
  • The Outlaw Run roller coaster in Branson, Missouri, features a track that is inclined at 84 ∘ below the horizontal and that span
    8·1 answer
  • How does alpha particles transfer energy to their surroundings?
    12·1 answer
  • While watching a planet orbiting a star, you notice that the doppler shift in the light of the star is becoming increasingly blu
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!