Ray creates an energy transfer diagram for a hair dryer. However, the diagram contains an error that could be corrected in sever
al different ways.
Which of these changes would make the diagram error-free? Check all that apply.
1. reducing the amount of thermal energy to 475 J
2. increasing the amount of electrical energy to 825 J
3. increasing the amount of sound energy to 225 J
4. reducing the amount of kinetic energy to 50 J
5. reducing the amount of electrical energy to 675 J
Which of these changes would make the diagram error-free? Check all that apply.
1. reducing the amount of thermal energy to 475 J
2. increasing the amount of electrical energy to 825 J
3. increasing the amount of sound energy to 225 J
4. reducing the amount of kinetic energy to 50 J
5. reducing the amount of electrical energy to 675 J
2 answers:
The correct answer is 1,2,4.
Option 1
Since the hair drier does not create any energy, the sum of the input energy has to be equal to the sum of the output energy. The input energy is 750J. In this case, sound energy is 150J and the kinetic energy is 125J. When the thermal energy is 475J, the output energy is,
In this case, the diagram is error free.
Option 2
In this option, the input energy is 825J .The sum of the input energy has to be equal to the output energy. The sum of the output energy is calculated as shown below,
In this situation, the diagram is error free because energy is conserved.
option 4
The input energy into this system is 750J. When the kinetic energy is reduced to 50J, the output energy of the drier is then,
In this case, energy conservation is not violated, the input energy is equal to the output energy.
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Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course : = 0.75 m/s²,
= 20 m,
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s,
= -1.15 m/s²,
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) × )
0 = 129.96 - 2.3
2.3 = 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √( ² +
² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
Answer:
v = 54.2 m / s
Explanation:
Let's use energy conservation for this problem.
Starting point Higher
Em₀ = U = m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ 2gh
Let's calculate
v = √ (2 9.8 150)
v = 54.2 m / s