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3 years ago

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A box is pulled 6 meters across the ground at a constant velocity by a horizontally applied force of 50 newtons. At the same tim

e, kinetic friction acts on the box as it slides so that it eventually stops when the pulling force stops. What is the magnitude of the frictional force? How much work does the frictional force do? What is the net work done on the box?

1 answer:

Vikentia [17]3 years ago

7 0

Answer:

(a) Friction force = 50 N

(b) Work done by friction = 300 j

(c) Net work done = 0 j

Explanation:

We have given that the box is pulled by 6 meter so d = 6 m

Force applied on the box F = 60 N

We have have given that velocity is constant so acceleration will be zero

So to applied force will be utilized in balancing the friction force

So friction force $F_{friction}=50N$

Work done by friction force $W_{friction}=F_{friction}\times d=50\times 6=300j$

Work done by applied force $W=F\times d=50\times 6=300j$

So net work done = 300-300 = 0 j

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Which data set has the largest standard deviation

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Answer:

<em>The data set marked as B has the largest standard deviation</em>

Explanation:

<u>Standard Deviation</u>

It's a number used to show how a set of measurements is spread out from the average value. A low standard deviation means that most of the values are close to the average. A high standard deviation means that the numbers are more spread out.

The formula for the standard deviation is

$\displaystyle \sigma=\sqrt{\frac{\sum (x_i-\mu)^2}{n}}$

Where $x_i$ is the value of each measurement, n is the number of elements in the set, and $\mu$ is the average or media of the values, defined as

$\displaystyle \mu=\frac{\sum x_i}{n}$

Let's analyze each set of data:

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The average is

$\displaystyle \mu=\frac{3+4+3+4+3+4+3}{7}=3.43$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2}{7}}$

$\sigma=0.5$

B.1,6,3,15,4,12,8

The average is

$\displaystyle \mu=\frac{1+6+3+15+4+12+8}{7}=7$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(1-7)^2+(6-7)^2+(3-7)^2+(15-7)^2+(4-7)^2+(12-7)^2+(8-7)^2}{7}}$

$\sigma=4.7$

C. 20, 21,23,19,19,20,20

The average is

$\displaystyle \mu=\frac{20+21+23+19+19+20+20}{7}=20.29$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(20-20.29)^2+(21-20.29)^2+(23-20.29)^2+(19-20.29)^2+(19-20.29)^2+(20-20.29)^2+(20-20.29)^2}{7}}$

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D.12,14,13,14,12,13,12

The average is

$\displaystyle \mu=\frac{12+14+13+14+12+13+12}{7}=12.86$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(12-12.86)^2+(14-12.86)^2+(13-12.86)^2+(14-12.86)^2+(12-12.86)^2+(13-12.86)^2+(12-12.86)^2}{7}}$

$\sigma=0.8$

We can see the data set marked as B has the largest standard deviation

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3 years ago

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Substituting into the equation and re-arranging it, we can find the corresponding frequency:

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And using $f=5.45 \cdot 10^{14} Hz$ as we found in the previous part, we can find the period of this wave:

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Can a vector have a component greater than its magnitude

galina1969 [7]

Answer:

NO

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Answer:

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Explanation:

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Read 2 more answers

A stone that is thrown vertically upwards was having a velocity of 15m/s after reaches 2/3 of its maximum height. What is the ma

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<em>The correct answer is option</em><em> B.</em> The maximum height that can be reached by the stone is determined as 11.5 m.

<h3>Maximum height attained by the stone </h3>

The maximum height attained by the stone when it is a 2/3 of its total height is calculated as follows;

v² = u² - 2gh

where;

v is final velocity at maximum height, v = 0
u is initial velocity
g is acceleration due to gravity

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (15²)/(2 x 9.8)

h = 11.48 m

h = 11.5 m

Thus, the maximum height that can be reached by the stone is determined as 11.5 m

Learn more about maximum height here: brainly.com/question/12446886

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