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saul85 [17]
3 years ago
9

Explain all the energy conversions that take place while using a cell phone.

Physics
1 answer:
zhuklara [117]3 years ago
8 0

Answer:

#See solution for details.

Explanation:

-Chemical energy in the battery is converted into Electrical Energy which powers up the phone.

-The electrical energy is then converted to Light Energy when the phone is powered up, this is seen through the lightening up of the phone screen.

-During phone calls, the electrical energy is further converted to Sound Energy to allow for transmission of audio signals.

- As we continue to use the phone, the electrical energy is converted into heat energy which we feel due to an overheating battery.

-The cycle then repeats itself again whenever a phone is charged.

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If a ball starts with the velocity of 2.0m/s and accelerates with a constant rate of 0.50m/s squared for 2.5s what is the balls
baherus [9]

Acceleration can be any change in speed, increasing or decreasing.
You haven't said whether the ball is speeding up or slowing down.

If its acceleration is positive ... speed is increasing ... then in 2.5 seconds,
it GAINS (0.5 m/s² x 2.5 sec) = 2.5 m/s of speed.  Added to its initial
speed of 2.0 m/s, it ends up moving at 4.5 m/s.

If its acceleration is negative ... speed is decreasing ... then in 2.5 seconds,
it LOSES (0.5 m/s² x 2.5 sec) = 2.5 m/s of speed.  Added to its initial
speed of 2.0 m/s, it ends up moving at  -0.5 m/s.  That means that it ends up
moving in the opposite direction compared to its direction at the beginning of
the change.

7 0
3 years ago
_____________ circular motion occurs when an object is traveling with constant speed in a circle.
emmasim [6.3K]
Your answer is ''Uniform''.

Hope this helps :)
7 0
3 years ago
A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then
Paha777 [63]

Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

6 0
3 years ago
A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l
Lelechka [254]

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>
  • Consider a body of mass m kept on a weighing machine in a lift.
  • The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
  • The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
  • Here we have given with the actual weight of the body as 100lbs.
  • This 100lb child is standing on the scale or the weighing machine, when it is riding .
  • During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
  • There is also<em> mg </em>downwards and a normal reaction in the upward direction.
  • when we equate both the upward force and downward force, we get,

                             ma=mg-N\\N=mg-ma    i.e. during riding the scale reads a weight less than that of actual weight.

  • When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

#SPJ4

7 0
1 year ago
Which statements describe the book and the forces acting on it? Check all that apply. The forces are balanced. The forces are un
dedylja [7]

Answer:

2.The forces are unbalanced.

5.The net force is to the right.

6.The book is moving to the right.

Explanation:

correct on edge :)

5 0
3 years ago
Read 2 more answers
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