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3 years ago

Which of the following is an example of qualitative data

1 answer:

3 0

Qualitative data is easy to explain. You either agree or disagree in writing. Wrong. You agree, disagree, or qualify. Meaning agree to somewhat extend, but not entirely. So you can agree that people should walk instead of talk in the progressive manner, but not in the impulsive manner.

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Sound energy cannot travel through A vacuum,a wooden table,polluted air,pond water

FromTheMoon [43]

Sound energy cannot travel through a vacuum.

Explanation:

Waves are periodic disturbance of the space, which travel carrying energy but not matter.

There are two types of waves:

Mechanical waves: mechanical waves propagate through the vibrations of the particles in a medium. Examples of mechanical waves are sound waves.
Electromagnetic waves: these waves consist of periodic oscillations of electric and magnetic fields, perpendicular to each other. These waves do not need a medium to propagate, so they can also travel in a vacuum.

In this problem, we are analyzing sound energy, which is the energy carried by sound waves. Sound waves are mechanical waves, so they need a medium to propagate: therefore, they cannot travel through a vacuum, since there is no medium.

So, sound energy cannot travel through a vacuum.

Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

6 0

3 years ago

A mass of 0.40 kg is suspended on a spring which then stretches 10 cm. The mass is then removed and a second mass is placed on t

Vanyuwa [196]

Answer:

x' = 1.01 m

Explanation:

given,

mass suspended on the spring, m = 0.40 Kg

stretches to distance, x = 10 cm = 0. 1 m

now,

we know

m g = k x

where k is spring constant

0.4 x 9.8 = k x 0.1

k = 39.2 N/m

now, when second mass is attached to the spring work is equal to 20 J

work done by the spring is equal to

$W = \dfrac{1}{2}kx'^2$

$20= \dfrac{1}{2}\times 39.2\times x'^2$

x'² = 1.0204

x' = 1.01 m

hence, the spring is stretched to 1.01 m from the second mass.

7 0

3 years ago

True or false: All of the rivers in the United States are closely monitored For pollution levels

Nutka1998 [239]

All of the rivers in the United States are not monitored closely for pollution levels.

The statement in the question is False

<h3>Explanation:</h3>

Not all rivers in the United States are closely monitored for their pollution levels. There are volunteers who take responsibility of checking water bodies such as lakes, rivers, streams etc. They take the next step required after finding a water body contaminated.

But to say that all the rivers receive the same amount of attention and action is incorrect. To ensure that all the rivers are clean and pollution free, it would require a lot of man power in the said department which would collect the water samples from all these rivers and send it to labs for checking the pollution level of the particular river. It can be said that the rivers which are situated beside an industrial site, close to a town, will be the priority for checking pollution levels as it would directly cause harm to a small set of population if the toxicity is at an increased level.

3 0

2 years ago

An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef-

maw [93]

Answer:

Power requirement P for the banner is found to be 30.62 W
Power requirement P for the solid flat plate is found to be 653.225 W
Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The Cd for banner was given, whereas the Cd for a flat plate is generally found to be around 1.28 which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
Power requirement P for the smooth spherical balloon was found to be 40.08 W

Explanation:

First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:

v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
The power can be determined this equation: P = F . v, where F represents the drag-force that we will need to determine and v represents the velocity of the airplane
The equation to determine drag-force is: $F = 1/2 * d * C_d * A$

In the drag-force equation Cd represents the co-efficient of drag and A represents the frontal area of the banner/plate/balloon (the object being towed)

Frontal area A of the banner is : 25 x 0.8 = 20 m^2

Part a) We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = 1/2 * 0.06* 1.225 * 20 = 0.735 N. Now to find the power P we will use P = F . v i.e. 0.735 * 41.66 = 30.62 W

Part b) For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28

Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = 1/2 * 1.28 * 1.225 * 20 = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = 653.225 W

Part c) The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption

Part d) The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : pi* r^2 (r = d / 2). Now using the same method as above:

Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N

P = 0.962 * 41.66 = 40.08 W

4 0

3 years ago

A box is against a wall. A person pushes on the box, but the box does not move. Is this

Serhud [2]

Answer:

it is a force

Explanation:

a force is a push or a pull

7 0

2 years ago

Read 2 more answers