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2 years ago

11

At a wedding reception, you notice a child who looks like their mass is about 25 kg running across the dance floor then sliding

on their knees until they stop. If the coefficient of kinetic friction between the child's pants and the floor is 0.15, what is the friction force acting on them as they slide?

2 answers:

Step2247 [10]2 years ago

7 0

Answer: i think the answer is 25.0 and divided by 0.15

Explanation:

Vitek1552 [10]2 years ago

4 0

Mass of the object m = 25 kg

Coefficient of friction Uk = 0.15

Frictional force Ff = Uk x F => Ff = Uk x m x g

Ff = 0.15 x 25 x 9.8

Frictional Force Ff = 36.75 N

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B, he has to apply less force to lift the box because the mass has decreased.

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If m is decreased by half, then f decreases as well. It takes less force to push a lighter object across the floor.

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3 years ago

What is the total current in a parallel circuit containing a 12-V battery, a 2 Ω resistor, and a 4 Ω resistor?

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Answer:

V=IR

since the circuit is parallel both resistor have same voltage and different current value according to OHMs law.

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2 years ago

The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p

Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

$4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$3.081x10^{16}m$

(c) How many meters in a light-year?

$9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$63325AU$

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

Since p is small it can be represent as:

$p(rad) = \frac{1AU}{d}$ (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

$p(rad) = \frac{p('')}{2.06x10^5}$ (2)

Then, equation 2 can be replace in equation 1:

$\frac{p('')}{2.06x10^5} = \frac{1AU}{d}$

$\frac{d}{1AU} = \frac{2.06x10^5}{p('')}$ (3)

From equation 3 it can be see that $1pc = 2.06x10^5 AU$

<em>a) How many parsecs are there in one astronomical unit? </em>

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

<em>(b) How many meters are in a parsec? </em>

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

<em>(c) How many meters in a light-year? </em>

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

$x = (3x10^{8}m/s)(31536000s)$

$x = 9.46x10^{15}m$

<em>(d) How many astronomical units in a light-year?</em>

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

<em>(e) How many light-years in a parsec?</em>

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

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3 years ago

If jack was traveling north 120 miles and it took him 3 hours to get there. what is the velocity that jack was traveling? is thi

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Answers:

40 mp/h; Vector

Reason:

120/3 is 40 miles per hour.

Velocity is a vector measurement.

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3 years ago

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Answer:

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2 years ago