After the two legs, the helicopter’s horizontal distance from the point of take off is 979 m
For the first leg,
d = 294 m
θ = 59.4°
[tex]d_{x}[/tex] = d cos θ
[tex]d_{x}[/tex] = 294 * cos 59.4°
[tex]d_{x}[/tex] = 147 m
For the second leg,
d = 849 m
θ = 10.5°
[tex]d_{x}[/tex] = d cos θ
[tex]d_{x}[/tex] = 849 * cos 10.5°
[tex]d_{x}[/tex] = 832 m
Total horizontal distance = [tex]d_{x}[/tex] ( 1st leg ) + [tex]d_{x}[/tex] ( 2nd leg )
Total horizontal distance = 147 + 832
Total horizontal distance = 979 m
Therefore, after the two legs, the helicopter’s horizontal distance from the point of take off is 979 m
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The figure shows a 100 W light bulb 1 meter away from my finger. If my finger tip has an area of 1 cm2 and if the wavelength of the light from the bulb is λ = 588 nm = 588 × 10−9 m, then show that the number of photons hitting my finger per second is about 1015γ/second.1 Watt of power is 1 Joule/second.Number of photons per second?
We are given the following information
Energy of bulb = 100 W = 100 Joules/second
Area of fingertip = 1 cm² = 0.0001 m²
Wavelength of light = 588×10⁻⁹ m
Number of photons per second = ?
Let us first convert the wavelength into energy
[tex]E=\frac{h\cdot c}{\lambda}[/tex]Where h is the plank's constant (6.626x10⁻³⁴J.s), c is the speed of light (3×10⁸ m/s) and λ is the wavelength.
[tex]\begin{gathered} E=\frac{6.626\times10^{-34}\cdot3\times10^8}{588\times10^{-9}} \\ E=3.3806\times10^{-19}\; \frac{J}{s} \end{gathered}[/tex]A child playing by the side of a well is throwing a stone vertically into the well with an initial velocity of 5 m/s. if the stone falls into the water 3 s later, find the height, h, of the point which the stone is thrown from the surface of the water. (g=10 m/s2)
a) 15m
b)30m
c)45m
d)60m
The height of the well is 30m
We are given that ,
The initial velocity of the stone = u = 5m/s
The stone fall into well in time = t = 3s
The acceleration due to gravity = g = 10 m/s²
To get the height of the stone which is thrown from the surface of the water in the well by equation,
h = ut + 1/2 gt²
Where, u is initial velocity in m/s, h is height in m, t is time in secs, g is acceleration due to gravity which is vertically upward direction i.e. (-10m/s²)
h = (5m/s) (3s) + 1/2 (-10m/s²)(3s)²
h = (15m) - (5m/s²)(9s²)
h = -30 m
Therefore, the height of the stone which is thrown upward is 30 m . And here, we can neglect the negative sign due to height is never be in negative.
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Which is increased when the string of a stringed instrument is tightened?1) timbre2) pitch3) wavelength4) loudness
The pitch of a note is the same as its frequency.
Remember that te timbre depends on the shape and other properties of the vibrating object. The loudness is related to the amplitude of the wave, and the wavelength of sound is inversely proportional to the frequency.
On the other hand, the frequency of the sound of a string is proportional to the square root of the tension in the string. Then, if the tension increases (if the string is tightened), the frequency will also increase (and if the frequency increases, the wavelength decreases).
Therefore, the parameter that increases when a string of a stringed instrument is tightened, is the pitch.
Therefore, the answer is:
[tex]2)\text{ Pitch}[/tex]If a bus is traveling at 12m/s and a passenger on the bus is walking to the back of the bus at a velocity of 5m/s, what is the relative velocity of the passenger relative to the ground?
The relative velocity is 17m/s.
The relative velocity of the passenger relative to the ground can be found by applying the concept of relative motion.
speed of bus (vb)=12m/s
speed of passenger inside the bus(vp)= 5m/s opposite to the speed of bus
speed of passenger relative to the ground = v
v= vb+vp
v= 12+(-5), since passenger is traveling in opposite direction
v=7m/
Therefore, the velocity of passenger relative to the ground is 7m/s.
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The hydrogen-to-helium ratio is an important piece of evidence for the Big Bang Theory. What is the minimum hydrogen-helium ratio observed?A. 4:1B. 3:1C. 2:1
Correct answer is option B.
Answer:
3:1
Explanation:
What would the separation between two identical objects, one carrying
2 C
of positive charge and the other
2 C
of negative charge, have to be if the electrical force on each was precisely
2 N?
Please Help
The distance between the two charges is 134,164.1 m.
What is the distance between the two identical charges?
The distance between the two identical charges is determined by applying Coulomb's law as shown below.
F = kq²/r²
where;
K is Coulomb's constantq is the magnitude of the chargesr is the distance between the chargesF is the electric force between the two chargesr = √(kq²/F)
r = √(9 x 10⁹ x 2²) / 2)
r = 134,164.1 m
Thus, the distance between the two charges is determined by applying Coulomb's law.
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A 77 kg student traveling in a car with a constant velocity has a kinetic energy of 1.7 104 J. What is the speedometer reading of the car in km/h? answer with:_____km/h
ANSWER
75.65 km/h
EXPLANATION
Given:
• The student's mass, m = 77 kg
,• The kinetic energy of the student in the car, KE = 1.7 x 10⁴ J
Find:
• The speed read in the speedometer of the car, which is the speed of the student, v (in km/h)
The kinetic energy of an object with mass m, traveling at a speed v, is,
[tex]KE=\frac{1}{2}mv^2[/tex]Solving for v,
[tex]v=\sqrt{\frac{2KE}{m}}[/tex]Replace the known values and solve,
[tex]v=\sqrt{\frac{2\cdot1.7\cdot10^4J}{77kg}}\approx21.013m/s[/tex]Note that because the kinetic energy is given in Joules - which is equivalent to kg*m²/s², the speed we found is in m/s. Now, knowing that there are 3600 seconds in 1 hour and that 1 km is equivalent to 1000 m, we can convert this to km/s,
[tex]v=21.013\frac{m}{s}\cdot\frac{3600s}{1h}\cdot\frac{1km}{1000m}\approx75.65km/h[/tex]Hence, the speedometer reading of the car is 75.65 km/h, rounded to the nearest hundredth.
The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208 m is mounted on the same rotating shaft as the second pulley.
Answer: It should be the answer beginning like this
The linear spread is that
The radial acceleration of locations along the blade's outer edge is approximately 17580 [tex]m/s^2[/tex].
What is Radial acceleration?Radial acceleration describes the acceleration of an object travelling on a circular path towards the circle's center. It can be defined as the rate of change of tangential velocity with regard to time and is also known as centripetal acceleration.
Given:
A table saw's engine rotates at 3450 revolutions per minute.A V-belt connects a pulley that's attached to the motor shaft to a second pulley half the diameter.A 0.208 m circular saw blade is installed on the same rotating shaft as the second pulley.We know that the motor is rotating at 3450 rev/min. One revolution is equal to 2π radians, so we can convert the motor speed to radians per minute:
ω₁ = (3450 rev/min) x (2π rad/rev) = 21675π rad/min
The second pulley is half the diameter of the first pulley, so its angular speed, ω₂, is twice that of ω₁:
ω₂ = 2ω₁ = 43350π rad/min
The circular saw blade is mounted on the same shaft as the second pulley, so it also rotates at the same angular speed:
ω = ω₂ = 43350π rad/min
We can now calculate the linear speed of the small piece of wood moving at the same rate as the rim of the circular saw blade, indicated by v. The circumference of the circle is supplied by the rim of the circular saw blade:
C = πd = π(0.208 m) = 0.6548 m
The linear speed of the little piece of wood is equal to the tangential speed of the circular saw blade's rim:
v = ωr
where r is the circular saw blade's radius, given by half its diameter:
r = d/2 = 0.208/2 = 0.104 m
By substituting the values, we obtain:
v = r = (43350 rad/min) x (0.104 m) x (1/60) = approx. 23.0 m/s
As a result, the linear speed of the little piece of wood moving at the same rate as the rim of the circular saw blade is about 23.0 m/s.
Next, compute the radial acceleration of locations on the blade's outer edge, represented by. The radial acceleration is calculated as follows:
α = rω²
By substituting the values, we obtain:
r2 = (0.104 m) x (43350 rad/min)2 x (1/602) = 17580 m/s2 (approximate)
Therefore, the radial acceleration of points on the outer edge of the blade is approximately 17580 m/s². This high radial acceleration explains why sawdust doesn't stick to the teeth of the saw blade.
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Your question is incomplete, most probably the complete question is:
The motor of a table saw is rotating at 3450{\rm rev/min}. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208{\rm m}is mounted on the same rotating shaft as the second pulley.
The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed?
v =_______________________ m/s
Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't stick to its teeth.
\alpha=______________________m/s2
Hello, I am having a hard time understanding this question, is it possible for you to help me?
Answer: h = 14 m
Explanation:
From the information given,
weight of rocket = 10N
Recall, weight = mg
Thus, mg = 10
where
m = mass of object
g = acceleration due to gravity and its values is 9.8 m/s^2
At the exact top of the trajectory, the total mechanical energy = 140 J
Recall,
total mechanical energy = kinetic energy + potential energy
At the exact top, kinetic energy = 0
Thus,
140 = potential energy
Recall, potential energy = mgh
where h is the height of the object(in this case, h is the height at the exact top and its value is 140J). Thus,
140 = mgh
Substituting mg = 10, we have
140 = 10h
h = 140/10
h = 14 m
What the difference between velocity and speed
Answer:
Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.
Which of the following is an example of Newton's third law of motion?A. A skydiver slows down when her parachute opens.B. A grocery cart moves forward when it is pushed.C. A cannon recoils backwards when it is fired.D. A rolling rock slows down due to friction.
Explanation:
The third law of Newton says that when an object exerts a force on a second object, the first object experiences an equal and opposite force that is exerted by the second object.
So, the example that shows this law is:
C. A cannon recoils backward when it is fired.
Because the cannon e
Two points ___________ create a line.A. sometimesB. neverC. alwaysD. not enough information
According to the Euclidean Postulates, a straight line segment can be drawn joining any two points.
Therefore, the answer is:
[tex]C)\text{ Always}[/tex]A plane's average speed between 2 cities is 800 km/hr. If the trip takes 4.5 hours howfar does the plane fly?
Considering the average velocity can be written as:
[tex]v=\frac{\Delta s}{\Delta t}[/tex]We can isolate the distance, and we get:
[tex]\Delta s=v*\Delta t\rightarrow\Delta s=800*4.5=3600km[/tex]Then, the distance the plane flew is 3600km
Neptune circles the Sun at a distance of 4.50 × 1012 m once every 164 years. Saturn circles the Sun at a distance of 1.43 × 1012 m. What is the orbital period of Saturn?
The orbital time period of the Saturn is 29.6 years
We are given that,
Distance from Sun to Saturn is = a = 1.43 × 10¹²
The mass of the Sun is = M =1.99 × 10³⁰kg
The Gravitational constant = G = 6.67 × 10⁻¹¹N-m²kg⁻²
To find the orbital period of Saturn we can use the equation ,
[tex]T^{2} = \frac{4\pi }{GM}a^{3}[/tex]
Where, T is the orbital time period of the of the Saturn , M is the mass of the sun , G is the gravitational constant.
Therefore, after putting the value in above equation we can get,
[tex]T^{2} = \frac{4(\(3.14)^{2} }{(6.67*10)^{-11} )N-m^{2} kg^{-2}}(1.43*10^{12}) ^{3}m[/tex]
[tex]T^{2} = \sqrt{8.688*10^{17} } s[/tex]
[tex]T = 932094415.818s[/tex]
So that , from above to convert the orbital time period of Saturn from second into year i.e. above seconds divided by seconds (1 sec = 3.154 ×10⁷ Earth years)
Thus, the orbital time period can be ,
[tex]T = 29.6 years[/tex]
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PLEASE HELP :((
A particular cookie provides 54.0 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 103-kilogram weight 2.45-decimeters above the ground with an energy efficiency of 25%. How many repetitions of this exercise can the athlete do with the energy supplied from one of these cookies?
Energy effectiveness would be a term that refers to the proportion of input power over output. Power generation, as well as simply energy utilization, is the process of reducing the amount of energy used to produce goods and services.
Considering that,
A cookie contains 54.0 kcal of energy. An athlete utilizes the 54.0 kcal inside this cookie from input energy.
The following diagram illustrates the relationship among input as well as output energy:
Efficiency = output energy / input energy...(i)
Output energy = efficiency × input energy
By using equation (i)
⇒ output energy = 0.25 × 54 kcal = 13.5 kcal.
The lifting exercise has been performed n times for the output energy.
In terms of potential energy, such output energy could be written as follows:
Mass × gravity ×height.
So, energy per repetition = mgh = 103 kg × 9.8 m/ × (2.45 × 0.1m) = 247.303 J = 0.059 kcal.
So, Count of repetitions = sim of output energy / energy per repetition..(ii)
By using equation (ii)
Count of repetitions = 13.5 kcal / 0.059 kcal =229 repetitions.
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I’m confused about which electromagnetic waves have the lowest frequency
The eletromagnetic wave that has the highest frequency is the gamma rays. It also has the highest energy and shortest wavelengths.
On the other hand, the type of eletromagnetic wave that has the lowest frequency, lowest energy and longest wavelength is radio waves.
A total of 8.0 joules of work is done when a constant horizontal force of 2.0 newtons to the left is used to push a 3.0-kilogram box acrossa counter top. Determine the total horizontal distance the box moves.
ANSWER
4 meters
EXPLANATION
Given:
• Work = 8.0 J
,• Force = 2.0 N
,• Mass of the box = 3.0 kg
Unknown:
• Distance the box moves
Work is the product of the applied force in the direction of motion and the distance an object moves,
[tex]W=F\cdot d[/tex]In this case, a box is pushed horizontally with a force of 2N producing a work of 8J. Solve the equation above for d,
[tex]d=\frac{W}{N}=\frac{8.0J}{2.0N}=4m[/tex]Hence, the horizontal distance the box moves is 4 meters.
What is the kinetic energy of the ocean liner ?
In order to calculate the kinetic energy, we can use the formula:
[tex]E_k=\frac{mv^2}{2}[/tex]Where m is the mass in kg and v is the speed in m/s.
So, for the bullet, we have:
[tex]E_k=\frac{0.0014\cdot408^2}{2}=116.5\text{ J}[/tex]And for the liner, we have:
[tex]E_k=\frac{59000000\cdot12^2}{2}=4248000000\text{ J}[/tex]So the ocean liner has greater kinetic energy.
Use Newton’s Law of Universal Gravitation and Newton’s Second Law to
Find g = acceleration due to gravity
Show g is independent of mass
By equating the two forces, acceleration due to gravity g is obtained which is independent of mass
What is Newton’s Second Law ?The law state that the rate of change of momentum is directly proportional to the force applied.
From Newton’s Second Law, Force F = mg. And from Newton’s Law of Universal Gravitation, Force F = GMm/r²
Where m is the mass of the satellite or the body revolving round the earth.
Equate the two forces.
mg = GMm/r²
The two m cancelled out leaving
g = GM/r²
Where
g = Acceleration due to gravityM = Mass of the earthG = Universal gravitational constantr = Distance between them.Therefore, since the mass of the satellite m has cancelled out, acceleration due to gravity g is independent of mass.
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Given a DC battery of voltage, V = 4.00 V connected to a resistor R with a current I = 3.00 A through the resistor. What power is in this circuit? 15.5 W 12.0 W 39.4 W 45.5 W 8.88 W
12.0 W
Explanation
Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit. to find the power in the circuit we need to use the expression:
[tex]P=IV[/tex]where P is the powe I is the current and V is the voltage
Step 1
a)Let
[tex]\begin{gathered} I=\text{ 3 Amperes} \\ V=4.0\text{ volts} \end{gathered}[/tex]b) now,replace
[tex]\begin{gathered} P=IV \\ P=3\text{ A*4 V} \\ P=12\text{ W} \end{gathered}[/tex]therefore, the answer is
12.0 W
I hope this helps you
3. A rescuer jumped from an airship in the ocean 1.20 x 102 m above the water's surface. Whatwas her kinetic energy at the moment she was 30.0 m from the water's surface? What was herspeed at that moment assuming her mass is 60.0 kg?
Given data,
The initial velocity of the body is zero.
The distance travelled by the rescuer upto the height of 30 m from the water surface is,
[tex]\begin{gathered} S=102-30 \\ S=72\text{ m} \end{gathered}[/tex]The final velocity of the rescuer at the height 30 m is,
[tex]v^2-u^2=2gS[/tex]where g is the acceleration due to gravity.
Substituting the known values,
[tex]\begin{gathered} v^2=2\times9.8\times72 \\ v^2=1411.2 \\ v=37.6ms^2 \end{gathered}[/tex]Thus, the kinetic energy of the rescuer is,
[tex]K=\frac{1}{2}mv^2[/tex]Substituting the known values,
[tex]\begin{gathered} K=\frac{1}{2}\times60\times1411.2 \\ K=42336 \\ K=42.3\text{ KJ} \end{gathered}[/tex]Thus, the kinetic energy of the rescuer is 42.3 KJ and speed of the rescuer is 37.6 meter per second square.
Jason is pulling a box across the room. He is pulling with a force of 24 newtons and his arm is making a 44 angle with the horizontal, what is the horizontal component of the forcehe is pulling with
carts, bricks, and bands
4. Which two trials demonstrate the effect of a doubling of force upon the acceleration of a cart of constant mass?
a. Trials 2 and 4
b. Trials 2 and 6
c. Trials 4 and 7
d. Trials 6 and 7
A. The two trials that demonstrate the effect of a doubling of force upon the acceleration of a cart of constant mass are Trials 2 and 4.
What is the applied force on an object?The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceAt a constant mass;
F₁/a₁ = F₂/a₂
When the force is doubled, the acceleration of the object is given as;
a₂ = F₂a₁/F₁
a₂ = (2F₁ x a₁) / F₁
a₂ = 2a₁
From the trials,
acceleration of trial 2 = 0.51 m/s²
acceleration of trial 4 = 1 m/s²
Thus, the two trials that demonstrate the effect of a doubling of force upon the acceleration of a cart of constant mass are Trials 2 and 4.
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carts, bricks, and bands
3. Which statement describes the effect of a doubling of force upon the acceleration of a cart of constant mass?
a. Doubling the force will cause the acceleration to be twice the original value.
b. Doubling the force will cause the acceleration to be one-half the original value.
c. Doubling the force will cause the acceleration to be four times the original value.
d. Doubling the force will cause the acceleration to be one-fourth the original value.
The statement that describes the effect of a doubling a force at a constant mass is "doubling the force will cause the acceleration to be twice the original value.
The correct answer is option A.
What is the applied force on an object?
The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceAt a constant mass;
F₁/a₁ = F₂/a₂
When the force is doubled, the acceleration of the object is given as;
a₂ = F₂a₁/F₁
a₂ = (2F₁ x a₁) / F₁
a₂ = 2a₁
Thus, when the force on the cart is doubled and the mass is constant, the acceleration of the cart will double as well.
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Identify the kinematic equation which relates the velocity and time.
The kinematic equation which relates velocity and time is
[tex]v=v_0+at[/tex]As when the acceleratio
The picture below shows a person swinging a toy plane attached to a string in
uniform circular motion.
Which vector points in the direction of the centripetal acceleration of the
plane?
According to the image A vector points in the direction of the centripetal acceleration of the plane.
The correct option is C.
What is centripetal acceleration?Centripetal acceleration is a property of an object's motion along a circular path. Centripetal force refers to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.
Briefing:A person is seen in the image swinging a toy plane on a string in a smooth, circular motion. The velocity direction is tangent to the circular orbit and perpendicular to the direction of the position.
The direction of the velocity and the speed may both vary as an object travels in a circular orbit. The velocity's direction is continually shifting. While acceleration is constantly moving uniformly in a circular path toward the recent, tangent at each point indicates the direction of velocity at that place.
As a result, vector A is pointed in the plane's centripetal acceleration's direction.
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The complete question is -
The picture below shows a person swinging a toy plane attached to a string in uniform circular motion.
Which vector points in the direction of the centripetal acceleration of the
plane?
A-B
B-D
C-A
D-C
Please help me, it's asking me here how does static energy work?
Static energy is the energy due to motionless state of a particle. If a particle is at rest it possess static energy also known as potential energy. When some force is applied to the particle then the static energy gets converted into kinetic energy of the particle.
Pls quick will mark brainliest.
Lance is working in a library using a trolley to carry books. As he stacks books from the trolley on shelves, it gets easier to push or pull the trolley. Which part of Newton's laws of motion explains the increased ease of moving the trolley?
A. Objects at rest tend to stay at rest.
B. Objects in motion tend to stay in motion.
C. Every action has an equal and opposite reaction.
D. Larger objects require greater amounts of force to move.
Answer: D
Explanation:
Larger objects require greater amounts of force to move.
As he stacks books from the trolley on shelves, it gets easier to push or pull the trolley because the trolley gets lighter when the books are shifted from the trolley to the shelves. So, larger objects require greater amounts of force to move. Hence, option D is correct.
What is inertia?Inertia is a property of bodies that prevents them from moving or, if they are already moving, causes them to change the speed or direction of their motion. The inertia of a body is a passive quality that prohibits it from acting in any way besides opposing active agents like forces and torques.
A moving body continues to move not because of its inertia but rather because no force exists to stop it, alter its direction, or accelerate it.
According to the question, when there are books on the trolley they require a greater amount of force to move but when the books are shifted on the shelves it's easier to move the trolley. So inertia is proportional to mass.
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An installation consists of a 30-kVA, 3-phase transformer, a 480-volt primary, and a 240-volt secondary. Calculate the largest standard size circuit breaker permitted for primary-only protection to be used without applying Note 1 of Table 450.3(B).
Answer: 45 A
Explanation:
Primary only protection 3-phase
I = 3 phase kVA / ( 1.723 * V)
I = 30000 / ( 1.732 * 480 ) = 36.085 A
Table 450.3(B)
Currents of 9A or more column
primary only protection = 125%
Max OCPD pri = 125% of I = 1.25 * 36.085 = 45.11 A
Table 450.3(B) Note 1 does not apply, use next smaller Table 240.6(A)
Next smaller = 45 A
what is the maximum efficiency that a heat engine could have when operating between the normal boiling and freezing temperatures of water
The maximum efficiency that a heat engine could have when operating between the normal boiling and freezing temperatures of water is 26.8 %
η = ( [tex]T_{H}[/tex] - [tex]T_{C}[/tex] ) / [tex]T_{H}[/tex] * 100
η = Efficiency
[tex]T_{H}[/tex] = Hottest temperature
[tex]T_{C}[/tex] = Coldest temperature
Hottest temperature = Boiling point
Coldest temperature = Freezing point
[tex]T_{H}[/tex] = 100 °C = 373 K
[tex]T_{C}[/tex] = 0 °C = 272 K
η = ( 373 - 273 ) / 373 * 100
η = 100 / 373 * 100
η = 26.8 %
In a heat engine, the heat energy is converted into mechanical energy which will be used to do mechanical work like pushing a piston out from the cylinder.
Therefore, the maximum efficiency that a heat engine could have when operating between the normal boiling and freezing temperatures of water is 26.8 %
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