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2 years ago

Suppose you are measuring the height of a small child. What will determine the number if significant digits you record?

1 answer:

8 0

The number of significant digits of any measurement is determined by the instrument used for such measurement. For example, in this case, we have the height of a small child being measured. We can use a simple ruler for this, and we see that a ruler has ten divisions for 1 cm. This means that the ruler cannot measure beyond the size of 0.1 cm or 1 mm. Hence, when we report the height of the small child, we report it to one significant digit after the decimal place. As an example, if we measure a child's height to be 90 full cm divisions and 8 smaller divisions, we report it as 90.8 cm but not 90.83 or 90.86 cm.

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The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi

ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done = mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = $= \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})$

kinetic energy = $= \frac{1}{2}*4*(1.5^{2}-11^{2})$ = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

7 0

2 years ago

A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitu

Mumz [18]

Answer:

maximum amplitude = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

$T=\dfrac{2\pi}{\omega}$

$0.74=\dfrac{2\pi}{\omega}$

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude = 0.13 m

8 0

2 years ago

What conclusion can be derived by comparing the central tendencies of the two data sets?

zhannawk [14.2K]

The answer is B. I don’t think I need to explain this,
Mean is average, Mode is the most common number, and Median is the middle number when you put the numbers is numerical order from least to greatest

3 0

2 years ago

If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol

xenn [34]

Answer:

Average density of Sun is 1.3927 $\frac{g}{cm}$ .

Given:

Radius of Sun = 7.001 × $10^{5}$ km = 7.001 × $10^{10}$ cm

Mass of Sun = 2 × $10^{30}$ kg = 2 × $10^{33}$ g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Solution:

Density of Sun is given by,

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Volume of Sun = $\frac{4}{3} \pi r^{3}$

Volume of Sun = $\frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}$

Volume of Sun = 1.436 × $10^{33}$ $cm^{3}$

Density of Sun = $\frac{ 2\times 10^{33} }{1.436 \times 10^{33} }$

Density of Sun = 1.3927 $\frac{g}{cm}$

Thus, Average density of Sun is 1.3927 $\frac{g}{cm}$ .

4 0

2 years ago

A student lifts a set of books off a table and places them in the upper shelf of a book case which is 2 meters above the table.

AfilCa [17]

The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle $\theta$ between the direction of the force and the direction of the displacement:
$W=Fd \cos \theta$
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so $\cos \theta = \cos 0=1$
Therefore, the work done is
$W=Fd=(5 N)(2 m)=10 J$

6 0

2 years ago

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