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bekas [8.4K]
2 years ago
5

Consider a heat engine that inputs 10 kJ of heat and outputs 5 kJ of work. What are the signs on the total heat transfer and tot

al work transfer
Physics
1 answer:
Oksi-84 [34.3K]2 years ago
5 0

Answer:

Total heat transfer is positive

Total work transfer is positive

Explanation:

The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.

Heat transfer to a system is positive and that transferred from the system is negative.

Also, work done by a system is positive while the work done on the system is negative.

Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)

Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).

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A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of fl
Korvikt [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is \frac{dT}{dt} = 1.016 ^oC/m

Explanation:

From the question we are told that

   The velocity field with which the bird is flying is  \vec V =  (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s

   The temperature of the room is  T(x, y, u) =  400 -0.4y -0.6z-0.2(5 - x)^2 \  ^o C

    The time considered is  t =  10 \  seconds

    The  distance that the bird flew is  x  =  1 m

 Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

      \frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 *  (5-x)] [-\frac{dx}{dt} ]

Here the negative sign in \frac{dx}{dt} is because of the negative sign that is attached to x in the equation

 So

       \frac{dT}{dt} = -0.4v_y  -0.6v_z -0.2[2 *  (5-x)][ -v_x]

From the given equation of velocity field

    v_x  =  0.6x

    v_y  =  0.2t

     v_z  =  -1.4

So

\frac{dT}{dt} = -0.4[0.2t]  -0.6[-1.4] -0.2[2 *  (5-x)][ -[0.6x]]    

substituting the given values of x and t

\frac{dT}{dt} = -0.4[0.2(10)]  -0.6[-1.4] -0.2[2 *  (5-1)][ -[0.61]]      

\frac{dT}{dt} = -0.8 +0.84 + 0.976  

\frac{dT}{dt} = 1.016 ^oC/m  

5 0
2 years ago
You need to design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a
Svetllana [295]

Answer:

Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram

Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely

used integrated circuits for creating clock pulses is called a 555 timer.  shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?

ANSWER : R1 = 144.3Ω,   R2 =  72.2 Ω

Explanation:

Frequency = 10 MHz

Time period = 1 / F =  0.1 <em>u </em>s

Duty cycle = 75% = 0.75

Duty cycle can be represented as :   Ton / T

Also: Ton = Th = 0.75 * 0.1 <em>u </em>s  = 75 <em>n</em> s

TL = T - Th = 100 <em>n</em>s - 75 <em>n</em> s = 25 <em>n</em> s

To find the value of R2 we use the equation for  time spent in the low (0 V) state

TL = R2*C*ln(2)

hence R2 = TL / ( C * In 2 )

c = 500 pF

Hence R2 = 25 / ( 500 pF * 0.693 )  = 72.2 Ω

To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,

Th = (R1 + R2)*C*ln(2)

  from the equation make R1 the subject of the formula

R1 =  (Th - ( R2 * C * In2 )) / (C * In 2)

R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )

R1 = ( 75 ns  - ( 25 ns ) / 500 pf * 0.693

     = 144.3Ω

8 0
2 years ago
A BMX bicycle rider takes off from a ramp at a point 2.4 m above the ground. The ramp is angled at 40 degrees from the horizonta
adoni [48]

Answer:

The BMX lands 5.4 m from the end of the ramp.

Explanation:

Hi there!

The position of the BMX is given by the position vector "r":

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t

x0 = initial horizontal position

v0 = initial velocity

α = jumping angle

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

Please, see the attached graphic for a better understanding of the situation. At final time, when the bicycle reaches the ground, the vector position will be "r final" (see figure). The y-component of the vector "r final" is - 2.4 m (placing the origin of the frame of reference at the jumping point). With that information, we can use the equation of the y-component of the vector "r" (see above) to calculate the time of flight. With that time, we can then obtain the x-component (rx in the figure) of the vector "r final". Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

-2.4 m = 0 m + 5.9 m/s · t · sin 40° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 5.9 m/s · t · sin 40° + 2.4 m

Solving the quadratic equation:

t = 1.2 s

Now, we can calculate the x-component of the vector "r final" that is the horizontal distance traveled by the bicycle:

x = x0 + v0 · t · cos α

x = 0 m + 5.9 m/s · 1.2 s · cos 40°

x = 5.4 m

The BMX lands 5.4 m from the end of the ramp.

Have a nice day!

8 0
2 years ago
Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is
GarryVolchara [31]

that would be given by

[email protected]

@ representing coefficient of kinetic friction.

thus 19.5/51.7 = 0.377

6 0
2 years ago
In which of the following color gradient types does the linear gradient gets mirrored on either side of the starting point?
Travka [436]

Answer: Reflected  

Explanation: Color Gradient refers to the variety of position dependent (as the vary with position) colors which are used to fill a region. Gradient is a mixture of two or more colors. This blend of colors adds depth to the design.

Types of color gradient are

  • Linear: The color blends in a straight from starting to ending point
  • Radial: The color transitions from starting point to ending a circular pattern.
  • Angle: It glides counterclockwise around starting point.
  • Diamond: It makes a diamond shape from starting point and its ending point is one corner of diamond.
  • Reflected: In this color gradient type symmetric linear gradient gets mirrored on either side of the starting point. This type of color gradient can be used to depict a lead pipe.

4 0
2 years ago
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