Consider a Carnot heat-engine cycle executed in a closed system using 0.025 kg of steam as the working fluid. It is known that the maximum absolute temperature in the cycle is twice the minimum absolute temperature, and the net work output of the cycle is 60 kJ. If the steam changes from saturated vapor to saturated liquid during heat rejection, determine the temperature of the steam during the heat rejection process. The temperature of the steam during the heat rejection process is

Answers

Answer 1

Answer:

The temperature of the steam during the heat rejection process is 42.5°C

Explanation:

Given the data in the question;

the maximum temperature T[tex]_H[/tex] in the cycle is twice the minimum absolute temperature T[tex]_L[/tex] in the cycle

T[tex]_H[/tex]  = 0.5T[tex]_L[/tex]

now, we find the efficiency of the Carnot cycle engine

η[tex]_{th[/tex] = 1 - T[tex]_L[/tex]/T[tex]_H[/tex]

η[tex]_{th[/tex] = 1 - T[tex]_L[/tex]/0.5T[tex]_L[/tex]

η[tex]_{th[/tex] = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η[tex]_{th[/tex] = 1 - W[tex]_{net[/tex]/Q[tex]_H[/tex]

where W[tex]_{net[/tex] is net work done, Q[tex]_H[/tex] is is the heat supplied

we substitute

0.5 = 60 / Q[tex]_H[/tex]

Q[tex]_H[/tex] = 60 / 0.5

Q[tex]_H[/tex] = 120 kJ

Now, we apply the first law of thermodynamics to the system

W[tex]_{net[/tex] = Q[tex]_H[/tex] - Q[tex]_L[/tex]

60 = 120 -  Q[tex]_L[/tex]

Q[tex]_L[/tex] = 60 kJ

now, the amount of heat rejection per kg of steam is;

q[tex]_L[/tex] = Q[tex]_L[/tex]/m

we substitute

q[tex]_L[/tex] = 60/0.025

q[tex]_L[/tex] = 2400 kJ/kg

which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)

q[tex]_L[/tex] = h[tex]_{fg[/tex] = 2400 kJ/kg

now, at  h[tex]_{fg[/tex]  = 2400 kJ/kg from saturated water tables;

T[tex]_L[/tex] = 40 + ( 45 - 40 ) ( [tex]\frac{2400-2406.0}{2394.0-2406.0}\\}[/tex] )

T[tex]_L[/tex] = 40 + (5) × (0.5)

T[tex]_L[/tex] = 40 + 2.5

T[tex]_L[/tex] = 42.5°C  

Therefore, The temperature of the steam during the heat rejection process is 42.5°C  


Related Questions

PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE USE TRUTHFUL ANSWERS PLEASE



Which best explains why species living in Australia are found nowhere else on Earth? This is an example of Geologic Evolution.

A.
Australia has an ecosystem different from any other area on Earth.

B.
Humans have genetically altered many Australian species in laboratories.

C.
Australian species were genetically altered after a comet hit the landmass.

D.
Australia separated from other continents and species there evolved independently.

Answers

Australia separated from other continents and species there evolved independently

In a magnetized object, most of the domains point
a. in the same direction.
b. in an east-west direction.
c. in different directions.
d. perpendicular to one another.

Answers

the answer for this question would the letter a

A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe very large accelerations.
Part A) What is the acceleration at the end of a test tubethat is 10 cm from the axis of rotation?
Part B) For comparison, what is the magnitude of theacceleration a test tube would experience if dropped from a heightof 1.0 m and stopped in a 1.0-ms-long encounter with a hardfloor?

Answers

Answer:

A)  a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

               a_c = v² / r

the angular and linear variables are related

              v = w r

we substitute

               a_c = w² r

let's reduce the magnitudes to the SI system

              w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

               r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

              a_c = 418.88² 0.1

               a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

                v² = v₀² - 2g (y- y₀)

                v² = 0 - 2 9.8 (0 + 1)

                v =√19.6

                v = -4.427 m / s

now let's look for the applied steel to stop the test tube

                v_f = v + a t

                0 = v + at

                a = -v / t

                a = 4.427 / 0.001

                a = 4.43 10³ m / s²

Let's assume raspberries are 10 wt% protein solids and the remainder water. When making jam, raspberries are crushed and mixed with sugar, in a 45:55 berry to sugar ratio, by mass. Afterward, the mixture is heated, boiling off water until the remaining mixture is 0.4 weight fraction water, resulting in the final product, jam. How much water, in kilograms, is boiled off per kilogram of raspberries processed

Answers

Answer:

The mass of water boiled off is [tex]0.0 \overline{185}[/tex] kg

Explanation:

The given percentage by weight of protein solids in raspberries = 10 weight%  

The ratio of sugar to raspberries in ja-m = 45:55

The mass of the mixture after boiling = 0.4 weight fraction water

Let 's' represent the mass of sugar in the mixture, and let 'r' represent the mass of raspberry

The mass of raspberry, r = 1 kg

The percentage by weight of water in raspberry = 90 weight %

The mass of water in 1 kg of raspberry =  90/100 × 1 kg = 0.9 kg

The ratio of the mass of sugar to the mass of raspberry in jam = r/s = 45/55

∴ s = 1 kg × 55/45 = 11/9 kg

The mass of the mixture before boiling = 1 kg + 11/9 kg = 20/9 kg

The weight fraction of water in the remaining mixture after boiling = 0.4 weight fraction

Let 'w' represent the mass of water boiled off, we have;

(0.9 - w)/(20/9 - w) = 0.4

(0.9 - w) = 0.4 × (20/9 - w)

0.9 - w = 8/9 - 0.4·w

9/10 - 8/9 = w - 0.4·w = 0.6·w = (6/10)·w

(81 - 80)/(90) = (6/10)·w

1/90 = (6/10)·w

w = ((10/6) × 1/90) = 1/54

w = 1/54  

The mass of water boiled off, w = (1/54) kg = [tex]0.0 \overline{185}[/tex] kg

plz help me with my career!!!
part one...

Answers

Answer:

#1 Yes

Explanation: #1: The rest of them are used mainley by farmers, and crops are used by common citizens in the world.

Question 1: Crops.

Question 2: Diagnostic Services.

Question 3: A cable company needs to lay new fiber optic cable to reach its customers across a large lake.

Question 4: A bachelor's degree in energy research.

Question 5: Environmental Resources.

If any of these answers are incorrect, please tell me, so I can fix my mistake. Thank you.

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.940 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor. m/s (b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2 (c) Calculate the force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Answers

Answer:A)u =4.295m/s  , B)a = 29.746m/s²   C) F=3,153N

Explanation:

Using the kinematic expression  

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

Given that he reaches a height of 0.940 m above the floor,

the final velocity  = 0

Here, acceleration due to gravity is acting in  opposite the initial direction of motion. So, a=-9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.940

- u² = 2 × - 9.81 × 0.920

- u² = -18.4428

cancelling the minus in both sides , we have that  

u² = 18.4428

u = √18.4428

u =4.295m/s

(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2

Using v² = u² + 2as

where u = initial speed of basketball player before lengthening = 0 m/s,

v = final speed of basketball player after lengthening =  4.295m/s,

a = acceleration while  straightening his legs

s = distance moved during lengthening = 0.310m

v² = u² + 2as  

 a = (v² - u²)/2s

a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)

a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)

a = (18.4428 m²/s²/(0.62 m)

a = 29.746m/s²

c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Force= mass x acceleration.

F = 106 kg X 29.746m/s²

 F = 3,153.076 rounded to  3,153N

A trough is 10 meters long, 1 meters wide, and 2 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 2 meters, and base, on top, of length 1 meters). The trough is full of water (density 1000kg/m3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top. (Note: Use g

Answers

Answer:

The amount of work required to empty the trough by pumping the water over the top is approximately 98,000 J

Explanation:

The length of the trough = 10 meters

The width of the through = 1 meter

The depth of the trough = 2 meters

The vertical cross section of the through = An isosceles triangle

The density of water in the through = 1000 kg/m³

Let 'x' represent the width of the water at a depth

x/y = 1/2

∴x = y/2

The volume of a layer of water, dV, is given as follows;

dV = 10 × y/2 × dy = 5·y·dy

The mass of the layer of water, m = ρ × dV

∴ m = 1000 kg/m³ × 5·y·dy m³ = 5,000·y·dy kg

The work done, W = m·g·h

Where;

h = The the depth of the trough from which water is pumped

g = The acceleration due to gravity ≈ 9.8 m/s²

[tex]\therefore \, W \approx \int\limits^2_0 {5,000 \times y \times 9.8 \, dy} = \left[24,500\cdot y^2 \right]^2_0 = 98,000[/tex]

The work done by the pump to pump all the water in the trough, over the top W ≈ 98,000 J

10. A person is about to kick a soccer ball. Consider the leg and foot to be a single rigid body and that it rotates about a fixed axis through the knee joint center. Immediately prior to impact the leg and foot are in the vertical direction and the distal end of the foot has an acceleration of 10 g in the horizontal direction. The muscle force vector makes an angle of 15 degrees with the vertical and has a moment arm of 5 cm from the knee joint center. Assume the person is 1.7 m tall and has a mass of 75 kg. Find the force in the muscle

Answers

Answer:

i don't know but i hope you get it right

Explanation

A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.

Answers

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2               ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1                  ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Fat Albert (the TV show character) runs up the stairs on Monday. On Tuesday, he walks up the same set of stairs. Which day did he do more work?

Answers

Answer:

Tuesday bc instead of running he/she was walking bc he/she might not have as much energy

Explanation:

What is an electric current?

a
a potential difference
b
a piece of wire
c
A flow of charge within a circuit
d
A light bulb
What is an electric current?

a
a potential difference
b
a piece of wire
c
A flow of charge within a circuit
d
A light bulb

Answers

Answer:

c

Explanation:

mark brainleyest plz if right

Fossil clues are one of the _____________ clues that support the theory of continental drift.
A. crust B. resource C. climate D. rock

Answers

Answer:

a

Explanation:

I think don't get mad if I'm wrong

Which one is it? Help ASAP

Answers

Answer:

extreme heat, because no physical damage can demagnetize a magnet

Explanation:

Answer:

the 3rd one

Explanation:

Part D Here is one last question as a final check on your understanding of your work for this problem, looking at this problem as an example of the Conservation of Energy. The action in this problem begins at location A , with the block resting against the uncompressed spring. The action ends at location B, with the block moving up the ramp at a measured speed of 7.35 m/s . From A to B, what has been the work done by non-conservative forces, and what has been the change in the mechanical energy of the block-Earth system (the ramp is a part of the Earth)

Answers

Answer:

The answer is "39.95 J".

Explanation:

Please find the complete question in the attached file.

[tex]\to W_{AC}=(\mu \ m \ g \ \cos \theta ) d[/tex]

            [tex]=(0.45 \times 1.60 \times 9.8 \times \cos 26^{\circ}) 6.30 \\\\=(7.056 \times \cos 26^{\circ}) 6.30 \\\\=6.34189079\times 6.30\\\\=39.95 \ J\\\\[/tex]

[tex]\therefore \\\\\bold{\Delta E =39.95 \ J}[/tex]

A cat dozes on a stationary merry-go-round, at a radius of 7.0 m from the center of the ride. The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.9 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding

Answers

Answer:

0.6

Explanation:

Given that :

Radius, R = 7m

Period, T = 6.9s

The Coefficient of static friction, μs can be obtained using the relation :

μs = v² / 2gR

Recall, v = 2πR/T

μs becomes ;

μs = (2πR/T)² / 2gR

μs = (4π²R² / T²) ÷ 2gR

μs = (4π²R² / T²) * 1/ 2gR

μs = 4π²R / T²g

μs = 4π²*7 / 6.9^2 * 9.8

μs = 28π² / 466.578

μs = 276.34892 / 466.578

μs = 0.5922887

μs = 0.6

In an effort to be the star of the half-time show, the majorette twirls a highly unusual baton made up of four mases fastened to the ends of light rods. Each rod is 1.0 m lone. Find the moment of inertia of the system about an axis perpendicular to the page and passing through the point where the rods cross.

Answers

Answer:

"0.25 kg-m²" is the appropriate answer.

Explanation:

The diagram of the question is missing. Find the attachment of the diagram below.

According to the diagram, the values are:

m₁ = 0.2

m₂ = 0.3

m₃ = 0.3

m₄ = 0.2

d₁ = d₂ = d₃ = d₄ = 0.5 m

As we know,

The moment of inertia is:

⇒  [tex]I=\Sigma M_id_i^2[/tex]

then,

⇒  [tex]I=m_1d_1^2+m_2d_2^2+m_3d_3^2+m_4d_4^2[/tex]

⇒     [tex]=d^2(m_1+m_2+m_3+m_4)[/tex]

On substituting the values, we get

⇒     [tex]=0.5^2\times (0.2+0.3+0.3+0.2)[/tex]

⇒     [tex]=0.25\times 1[/tex]

⇒     [tex]=0.25 \ Kg-m^2[/tex]

Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
at 6◦ C. How many grams of ice will melt? The heat of fusion of water is 333 kJ/kg and the
specific heat is 4190 J/kg · K.Immersive Reader

Answers

Answer:

7.55 g

Explanation:

Using the relation :

Δt = temperature change = (6° - 0°) = 6°

Q = quantity of heat

C = specific heat capacity = 4190 j/kg/k

1000 J = 1kJ

333 KJ = 333000 j

The quantity of ice that will melt ;

= 0.419 * 6 * 100 / 333000

= 2514000 / 333000

= 7.549 g

The mass of ice that will melt :

2.514 / 0.333

= 7.549 g

Based on the definition of weather, ____________, is the basic difference climate and weather.

a)precipitation

b)condensation

C)time


D)jet stream

Answers

Answer:

The answer is C. time. Please mark as brainliest.

HELP PLEASE DUE IN 3 MINUTES

Answers

Answer:

tectonic plate movement

Explanation:

To understand the relationship between the force and the potential energy changes associated with that force and to be able to calculate the changes in potential energy as definite integrals. Imagine that a conservative force field is defined in a certain region of space. Does this sound too abstract

Answers

Answer:

This does not sound too abstract

Explanation:

An example of an Imaginative conservative force field defined in terms of a certain region of space is : A gravitational field and it can be expressed as

[tex]Fx = - \frac{dU_{x} }{dx}[/tex]  and this is because the potential energy in a gravitational field is determined only by the initial and final position of the objects found with the Gravitational field

To determine the change in potential energy  ΔU we will integrate the above equation

hence :  ΔU = [tex]\int\limits^2_1 {F} \, ds[/tex]  

Hence this does not sound too abstract

The theorist who expanded the behavioristic perspective to include cognitive influences on personality was
Please select the best answer from the choices provided
O Sigmund Freud
O Albert Bandura
OB. F. Skinner
O Carl Jung

Answers

Answer:

It is actually Albert Bandura! Hope this helps<3

Explanation:

In contrast to Skinner's idea that the environment alone determines behavior, Bandura (1990) proposed the concept of reciprocal determinism, in which cognitive processes, behavior, and context all interact, each factor simultaneously influencing and being influenced by the others.

The theorist who expanded the behavioristic perspective to include cognitive influences on personality was Albert Bandura. So, the correct option is B.

What is Personality?

Personality is defined as a distinctive set of behavior, cognition, and emotional patterns that are formed by biological and environmental factors, and that change over time. This explained as the enduring characteristics and behavior which comprise a person's unique adjustment to life, including major traits, interests, drives, values, self-concept, abilities, and emotional patterns.

The four theory is also called as the proto-psychological theory which suggests that there are four fundamental personality types: sanguine, choleric, melancholic, and phlegmatic. Albert Bandura (1990) proposed the concept of reciprocal determinism in which cognitive processes, behavior, and context all interact, with each factor simultaneously affecting and being affected by the others.

Thus, the theorist who expanded the behavioristic perspective to include cognitive influences on personality was Albert Bandura. So, the correct option is B.

Learn more about Personality development, here:

https://brainly.com/question/1452486

#SPJ6

A system consists of a copper tank whose mass is 13 kg, 4 kg of liquid water, and an electrical resistor of negligible mass. The system is insulated on its outer surface. Initially, the temperature of the copper is 27oC and the temperature of the water is 50oC. The electrical resistor transfers 120 kJ of energy to the system. Eventually the system comes to equilibrium. Determine the final equilibrium temperature (in oC).

Answers

Answer:

T =  30.42°C

Explanation:

According to the conservation of energy principle:

[tex]Energy\ Given\ by\ Resistor = Heat\ Gain\ by\ Copper + Heat\ Gain\ by\ Water\\E = m_{c}C_{c}(T_{2c} - T_{1c}) + m_{w}C_{w}(T_{2w} - T_{1w})[/tex]

E = 120 KJ

mc = mass of copper = 13 kg

Cc = specific heat capacity of copper = 0.385 KJ/kg.°C

T2c = T2w = Final Equilibrium Temperature = T = ?

T1c = Initial Temperature of Copper = 27°C

T1w = Initial Temperature of Water = 50°C

mw = mass of water = 4 kg

Cw = specific heat capacity of water = 4.2 KJ/kg.°C

Therefore,

[tex]120\ KJ = (13\ kg)(0.385\ KJ/kg^oC)(T-27^oC) + (4\ kg)(4.2\ KJ/kg^oC)(T-50^oC)\\120\ KJ - 135.135\ KJ - 840\ KJ = (- 5.005T - 16.8 T)\ KJ/^oC\\T = \frac{-855.135\ KJ}{-28.105\ KJ/^oC}\\[/tex]

T =  30.42°C

A spring in a toy gun has a spring constant of 10 N/m and can be compressed 4 cm.
It is then used to shoot a 1 g ball out of the gun. Find the velocity of the ball as it
leaves the gun

Answers

F=ma
We know from this that a=10,000m/s2

V=at
X=vt


You end up with v^2=ax

Plug in 10,000 and 0.04 and solve for v =20m/s

Suppose you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 50 mi/hr with a constant 5-second time headway between them. All vehicles in lane 2 are traveling at 55 mi/hr with a constant 6-second time headway between them. All vehicles in lane 3 are traveling at 60 mi/hr with a constant 10-second time headway between them. You collected spot speed data for all vehicles as they crossed your observation point for 75 minutes. After 10 minutes, vehicles in lane 1 stopped arriving, and after 30 minutes, vehicles in lane 2 stopped arriving. What is the flow and density of the observed traffic stream

Answers

Answer:

lane 3    Ф   = 450 vehicles,  ρ = 0.1 vehicle / s

lane 2    Ф_{average} = 300 vehicles,  ρ _{average} = 6.66 10⁻² vehicles/s

lane 1      Ф_{average} = 300 vehicles,  ρ_{average} = 2.66 10⁻² vehicle/s

Explanation:

Before solving this exercise we must clarify the concepts the flow is defined as the occurrence of an event in a time interval, in this case the passage of a car through time

Flux Density is the flux between unit area or unit time

Let's start by calculating the calculation for lane 3

the flow.

Let's use a direct rule of proportions (rule three) if the number of vehicles per unit of time (t₀ = 10s), for the observation time how many vehicles passed in the observation time (t_total = 75 * 60 = 4500 s)

            Ф = 4500 s (1 vehicle / 10 s)

            Ф   = 450 vehicles

The flux density is the flux per unit area, in this case the area is not indicated, so we can define the flux density as the flux per unit of time.

           ρ = 450/4500

           ρ = 0.1 vehicle / s

Lane 2

we look for the flow

we can have separates the interval into two parts

* for the first t₁1 = 30 * 60 = 1800 s

            Ф₁ = 1800 s (1 vehicle / 6s)

            Ф₁ = 300 vehicles during t₁

* for the rest of the time t₂ = 4500-1800 = 2700 s

           Ф₂ = 0

the average density is the total number of vehicles between the total time

           #_ {vehicle} = 300 +0

           

            Ф_{average) = # _vehicle

            Ф_{average} = 300 vehicles in all time

             

The density is

            ρ 1 = fi1 / t1

            ρ1 = 300/1800

            ρ1 = 1.66 10-1 vehicles / s

the average density is

            ρ_{average} = [tex]\frac{\phi_1 + \phi_2}{ t_{total}}[/tex]

            ρ _{average} = (300 +0) / 4500

            ρ _{average} = 6.66 10⁻² vehicles / s

Lane 1

flow

* first time interval t₁ = 10 * 60 = 600 s

             Ф₁ = 600 s (1 vehicle / 5s)

             Ф₁ = 120 vehicles in interval t₁

* second interval t₂ = 4500-600 = 3900 s

              Ф2 = 0

 average flow

             Ф = Ф1 + Ф2

             Ф = 120 vehicles at all time

Density

* first interval

          ρ₁ = 120/600

          ρ₁ = 0.2 vehicles / s

* second interval

          ρ₂ = 0

average density

         ρ+{average} = 120/4500

         ρ_{average} = 2.66 10⁻² vehicle/s

1) A man leans against the wall and...
a) The man's shoulder pushes against the wall
b) The wall pushes against the man
c) Bricks in the wall push against each other
d) The ground pushes against the man, supporting him.

Answers

Answer:

I thinks its a, but its really about gravity im not sure

Explanation:

:)

Suppose you are standing in front of a flat mirror which is mounted to a vertical wall. For this problem you may suppose that your height is 1.70 m and your eyes are 12 cm below the top of your head. What is the smallest mirror that will still allow you to see the full length of your body

Answers

Answer:

The right approach is "0.85 m".

Explanation:

According to the question, the diagram will is provided below.

So that as per the diagram,

The values will be:

My height,

AO = 1.70 m

My eyes at,

AB = 12 cm

i.e.,

     = 0.12 m

As we can see, the point of incidence lies between the feet as well as the eyes, then

BO = 1.58 m

Now,

⇒ [tex]O'D = \frac{1.58}{2}[/tex]

           [tex]=0.79 \ m[/tex]

The point of incidence of the ray will be:

⇒ [tex]CO'=1.70-\frac{0.12}{2}[/tex]

           [tex]=1.70-0.06[/tex]

           [tex]=1.64 \ m[/tex]

Hence,

The smallest length of the mirror will be:

= [tex]CO'-O'D[/tex]

On substituting the values, we get

= [tex]1.64-0.79[/tex]

= [tex]0.85 \ m[/tex]

HELP PLEASE DUE IN 3 MINUTES

Answers

Answer:

Tectonic Plate Movement

Explanation:

Each continent and ocean sits on its own tectonic plate which floats on the Earths upper mantle. They move very little over time.

Answer:

tectonic plates movement

How much heat is needed to boil 5.30 kg of water at its boiling point?
Use Q = mass x latent heat of vaporization.
A. 1760 kJ B. 22.2 kJ C. 530 kJ D. 12,000kJ

Answers

Answer:

Required heat Q = 11,978 KJ

Explanation:

Given:

Mass = 5.3 kg

Latent heat of vaporization of water = 2,260 KJ / KG

Find:

Required heat Q

Computation:

Required heat Q = Mass x Latent heat of vaporization of water

Required heat Q = 5.3 x 2260

Required heat Q = 11,978 KJ

Required heat Q = 12,000 KJ (Approx.)

Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 lb and was traveling eastward. Car B weighs 11251125 lb and was traveling westward at 42.042.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.517.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.7500.750 . How fast (in miles per hour) was car A traveling just before the collision

Answers

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

          Wₐ = 1500 lb

          W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

             W = mg

             m = W / g

             mₐ = Wₐ / g

             m_b = W_b / g

             mₐ = 1500/32 = 46.875 slug

             m_b = 125/32 = 35,156 slug

Let's reduce to the english system

             v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

           p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

           p_f = (mₐ + m_b) v

the moment is preserved

           p₀ = p_f

           mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

           v = [tex]\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}[/tex]

we substitute the values

           v = [tex]\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6[/tex]

           v = 0.559 v₀ₐ - 26.40                  (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

              M = mₐ + m_b

              M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

              K₀ = ½ M v²

final point. When they stop

             K_f = 0

The work is

             W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

           N-W = 0

           N = W

the friction force has the expression

            fr = μ N

we substitute

            -μ W x = Kf - Ko

             

            -μ W x = 0 - ½ (W / g) v²

            v² = 2 μ g x  

            v = [tex]\sqrt{ 2 \ 0.750 \ 32 \ 17.5}[/tex]Ra (2 0.750 32 17.5  

            v = 28.98 ft / s

A toy projectile is fired vertically from the ground upward with a velocity of +29 meters per second. It arrives at its maximum altitude in 3.0 seconds. How high does the projectile go?

Answers

Answer:

[tex]\boxed{\text{\sf \Large 42 m}}[/tex]

Explanation:

Use height formula

[tex]\displaystyle \sf H=\frac{u^2 sin(\theta)^2}{2g}[/tex]

u is initial velocity

θ = 90° (fired vertically upward)

g is acceleration of gravity

[tex]\displaystyle \sf H=\frac{29^2 \times sin(90 )^2}{2 \times 10}=42.05[/tex]

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