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2 years ago

If she runs 3,223 meters each day and keeps this up for two weeks, how many miles will she have run

Physics

1 answer:

seropon [69]2 years ago

5 0

Answer:

28 miles

Explanation:

3,223 meters/day x 14 days = 45,122 meters

1 meter = 3.28 feet

45,122 meters = 148,000 feet

5280 feet = 1 mile

148,000 feet = 28 miles

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A _____ has to do with the direction of a force

Elenna [48]

Answer:

The "solid force"? ... The direction of the force always seems to be coming out of the solid surface. A direction which is perpendicular to the plane of a surface is said to be normal. The force that a solid surface exerts on anything in the normal direction is called the normal force.

Explanation:

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3 years ago

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38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

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2 years ago

If the average velocity during the athlete's walk back

goblinko [34]

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

$T=\frac{d}{1.50}$

From the question we are told

If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent is mathematically given as

T=\frac{d}{v}

Therefore

$T=\frac{d}{1.50}$

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

$T=\frac{d}{1.50}$

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2 years ago

g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe

olga_2 [115]

Answer:

The minimum compression is $x= 0.046m$

Explanation:

From the question we are told that

The mass of the block is $m_b = 1.45 kg$

The spring constant is $k = 860 N/m$

The coefficient of static friction is $\mu = 0.36$

For the the block not slip it mean the sum of forces acting on the horizontal axis is equal to the forces acting on the vertical axis

Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

$F_y = m_b*g$

And the force acting on the horizontal axis is force due to the spring which is mathematically represented as

$F_x = k *x * \mu$

where x is the minimum compression to keep the block from slipping

Now equating this two formulas and making x the subject

$x = \frac{m_b * g}{k * \mu}$

substituting values we have

$x = \frac{1.45 * 9.8}{860 *0.36}$

$x= 0.046m$

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3 years ago

What systems of the body are involved in preparing and eating a sandwich

jek_recluse [69]

<span>The systems of the body involved in preparing and eating a sandwich are :
</span>The skeletal, muscular, and digestive system. They are all directly involved in eating and preparing a sandwich.

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3 years ago