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3 years ago

What can be used to describe the quantitative data

1 answer:

6 0

<span>Quantitative data can be analyzed in a variety of different ways. In this section, you will learn about the most common quantitative analysis procedures that are used in small program evaluation.
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MEASUREMENT

lubasha [3.4K]

Answer:

You would need to type the numbers in the question or you could have added a picture but you didn't so there is no way to answer this question. Have a nice day!

5 0

2 years ago

In the standing waves experiment, the string has a mass of 31.2 g and a length of 0.7 m. The string is connected to a mechanical

mestny [16]

Answer:

linear density of the string = 4.46 × 10⁻⁴ kg/m

Explanation:

given,

mass of the string = 31.2 g

length of string = 0.7 m

linear density of the string = $\dfrac{mass\ of\ string}{length}$

linear density of the string = $\dfrac{31.2\times 10^{-3}\ kg}{0.7\ m}$

linear density of the string = 44.57 × 10⁻³ kg/m

linear density of the string = 4.46 × 10⁻⁴ kg/m

7 0

3 years ago

An energy storage system based on a flywheel (a rotating disk) can store a maximum of 3.7 MJ when the flywheel is rotating at 16

Likurg_2 [28]

The moment of inertia of the flywheel is 2.63 kg- $m^{2}$

It is given that,

The maximum energy stored on the flywheel is given as

E=3.7MJ= 3.7× $10^{6}$ J

Angular velocity of the flywheel is 16000 $\frac{rev}{min}$ = 1675.51 $\frac{rad}{sec}$

So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

E = $\frac{1}{2}$ $Iw^{2}$

By rearranging the equation:

I = $\frac{2E}{w_{2} }$

I = 2.63 kg- $m^{2}$

Thus the moment of inertia of the flywheel is 2.63 kg- $m^{2}$ .

Learn more about moment of inertia here;

brainly.com/question/13449336

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7 0

1 year ago

A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for

faltersainse [42]

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

5 0

3 years ago

Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (

Sauron [17]

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

2.9 kg (0, 3.2)

4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

$x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}$

$0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}$

$10 + 8.3 x = 0$

$x = -1.20 m$

Similarly for y direction we have

$y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}$

$0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}$

$9.28 + 8.3 x = 0$

$x = -1.12 m$

5 0

3 years ago