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umka2103 [35]
2 years ago
13

An object moves with a positive acceleration. Could the object be moving with increasing speed, decreasing speed or constant spe

ed
Physics
1 answer:
Masja [62]2 years ago
3 0

Answer:

Increasing speed.

Explanation:

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

In this scenario, an object moves with a positive acceleration. Thus, the object is moving with an increasing speed and as such it has acceleration in the same direction as its velocity with respect to time.

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5.54 Two kilograms of air within a piston–cylinder assembly WP execute a Carnot power cycle with maximum and minimum temper atur
Step2247 [10]

Answer:

A) 60%

B) p2 = 1237.2 kPa

   v2 = 0.348 m^3

C) w1-2 = w3-4 = 1615.5 kJ

   Q2-3 = 60 kJ

Explanation:

A) calculate thermal efficiency

  Л = 1 - \frac{Tl}{Th}  

where Tl = 300 k

            Th = 750 k

hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%

B) calculate the pressure and volume at the beginning of the isothermal expansion

calculate pressure ( P2 )  :

= P3v3 = mRT3  ----- (1)

v3 = 0.4m , mR = 2* 0.287, T3 = 750

hence P3 = 1076.25

next equation to determine P2

Qex = p3v3 ln( p2/p3 )

60 = 1076.25 * 0.4 ln(p2/p3)

hence ; P2 = 1237.2 kpa

calculate volume ( V2 )

p2v2 = p3v3

v2 = p3v3 / p2

   = (1076.25 * 0.4 ) /  1237.2  

  = 0.348 m^3

C) calculate the work and heat transfer for each four processes

work :

W1-2 = mCv( T2 - T1 )

        = 2*0.718 ( 750 - 300 ) = 1615.5 kJ

W3-4 = 1615.5 kJ

heat transfer

Q2-3 = W2-3 = 60KJ

Q3-4 = 0

D ) sketch of the cycle on p-V coordinates

attached below  

6 0
3 years ago
Describe each of Newton’s Laws of Motion in ice skating. What can you design/develop to improve ice skating?
denis23 [38]

Newton's three laws of motion can be used to describe the motion of the ice skating.

<h3>Newton's first law of motion</h3>

Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.

  • Based on this law, once the ice skating starts, it will continue endlessly unless external force stops it.

<h3>Newton's second law of motion</h3>

Newton's second law of motion states that the force applied to an object is directly proportional to the product of mass and acceleration of an object.

  • Based on this law, the force applied to the ice skating is equal to the product of mass and acceleration of the ice skating.

<h3>Newton's third law of motion</h3>

This law states that action and reaction are equal and opposite.

  • Based on this law, the force applied to the ice skating is equal in magnitude to the reaction of ice.

Learn more about Newton's law here: brainly.com/question/3999427

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2 years ago
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Answer:

Most of us have experienced that pivotal peak of pain, anger or frustration in which we want to scream “I hate my life.” Yet, the feeling that a dark cloud has specifically settled over us and our experiences can feel pretty isolating. The truth is, no matter how singled out or overwhelmed we feel, and no matter what area we are struggling in, we are not alone.  More than half of U.S. workers are unhappy with their job.  One in 10 Americans struggles with depression. All of us have moments of utter despair. Escaping from this hopeless-seeming state may feel impossible. Yet, in reality, we are not doomed, and we are not powerless.  No matter what our circumstances, we can all learn tools to help us emerge from the darkest moments in our lives.

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3 years ago
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Earth attracts a person with a gravitational force of 7.0 × 102 newtons. What is the magnitude of the force with which the indiv
Anuta_ua [19.1K]
This is a good time to review Newton's 3rd law of motion:
"For every action, there is an equal and opposite reaction."

Gravitational force always acts in pairs.
Whatever force the Earth attracts something with,
the thing attracts the Earth with exactly the same force.

If Earth attracts a person with a gravitational force of <span><span>7.0 × 10² </span>newtons,
the person attracts Earth with a gravitational force of 7.0 × 10² newtons.

Your weight on Earth is the same as the Earth's weight on you !
</span>
5 0
3 years ago
A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan
PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d)  V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.  

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

                                 y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

                                 55 = 0 + 0 + 0.5*(9.81)*t^2

                                 t = sqrt ( 55 * 2 / 9.81 )

                                 t =3.349 s

- Use the second equation of motion in x direction:

                                 x(f) = x(0) + V_x,i*t

                                 150 = 0 + V_x,i*3.349

                                  V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

                                 V_y,f = V_y,i + g*t

                                 V_y,f = 0 + 9.81*3.349

                                 V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

                                 V^2 = V_y,f^2 + V_x,i^2

                                 V = sqrt (32.85^2 + 44.8^2)

                                 V = 55.55 m/s

5 0
3 years ago
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