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Aloiza [94]
3 years ago
7

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

Physics
2 answers:
Anettt [7]3 years ago
4 0

The 1st, 2nd, and 4thoptions are right :)

Tasya [4]3 years ago
3 0

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,

475J+150J+125J=750J.

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,

125J+150J+550J=825J.

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,

50J+150J+550J=750J.

In this case, energy conservation is not violated, the input energy is equal to the output energy.


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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Charra [1.4K]

Power = Work / time

The work given here is 83J and the time it took to do 83J of work was 3s

So..

Power = 83J / 3s

Power = 27.67 W or 27.7 W

8 0
3 years ago
A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

-W_{x}+f_{s}=0

when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

When seeing the free-body diagram we can determine that:

W_{y}=mg cos \alpha

so we can substitute that in the sum of y-forces equation, so we get:

N=mg cos \alpha

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

mg cos \alpha \mu_{s}=mg sin \alpha

we can divide both sides of the equation into mg so we get:

cos \alpha \mu_{s}=sin \alpha

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for \mu_{s} we get:

\mu_{s}=\frac{sin \alpha}{cos \alpha}

when simplifying this we get that:

\mu_{s}=tan \alpha

now we can solve for the angle so we get:

\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

which yields:

α=19.29°

which rounds to:

α=19°

8 0
3 years ago
Using -10 m/s2 for acceleration due to gravity, what would be the total displacement of the object if it took 8 seconds before h
ASHA 777 [7]
If it starts from 0m/s...
s=?
u=0
a=-10
t=8
s=ut +1/2at^2
so s=(0×8)+ (0.5×-10×64)
s=0+(32×-10)
s=32×-10
s=-320metres
6 0
3 years ago
A 2.0-kg object moving at 5.0 m/s encounters a 30-Newton restive force
sveta [45]

The impulse experienced by the object is 3 N s.

<u>Explanation:</u>

Impulse is also termed as change in the momentum of the object. So, it is directly proportional to the force acting on the object and the time for which the force is acting on that object.

Thus, impulse experienced by an object is the product of force acting on the object for a given time period. So, it is the sudden influence of force on the given volume.

As the force is given as 30 N and the duration or the time is given as 0.1 seconds. Then, the impulse will be product of force with duration.

Impulse = Force × ΔTime = Force × Duration

Impulse = 30 × 0.1 = 3 N s.

Thus, the impulse experienced by the object is 3 N s.

6 0
3 years ago
Question 3 of 5
kap26 [50]

Answer:

Heat

Explanation:

The term heat refers to the conduction process which occurs when energy is passed between objects

8 0
3 years ago
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