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3 years ago

9

A person in car slides to the right when going through a very sharp left turn on the highway. Is that law of inertia or law of a

cceleration?

1 answer:

inessss [21]3 years ago

4 0

Answer:

Inertia :)

Explanation:

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HELP NOW PLEASE ANSWER THE QUESTION THE PICTURE IS ATTACHED BELOW

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Answer:

c

Explanation:

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3 years ago

A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to

sineoko [7]

The temperature of the water getting colder would cause the liquid in the thermometer to drop due to less heat being transferred from the water to the liquid, so the liquid molecules are closer than when they have high energy.

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4 years ago

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Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has a sma

Helga [31]

Answer:

<em>Which box has the greater acceleration?</em>

E. Box A

Explanation:

<em>The question is incomplete:</em>

<em>Which box has the greater acceleration?</em>

The bouyant force exerted by the water is equal in both boxes, because it depends on the volume displaced (that is the same for both boxes) and the density of the water.

But, the weight of each boxes is different, according to their density.

For the Box A the acceleration will be:

$m_aa_a=gV(\rho_w-\rho_a)\\\\\rho_aVa_a=gV(\rho_w-\rho_a)\\\\a_a=g\frac{(\rho_w-\rho_a)}{\rho_a}$

The same applies for the Box B:

$a_b=g\frac{(\rho_w-\rho_b)}{\rho_b}$

If we express the ratio of the accelerations, we have:

$a_a/a_b=\frac{(\rho_w-\rho_a)}{\rho_a}*\frac{\rho_b}{(\rho_w-\rho_b)}\\\\$

$a_a/a_b=\frac{(\rho_w-\rho_a)}{(\rho_w-\rho_b)} \frac{\rho_b}{\rho_a}$

We know that both densities are lower than water, because they accelerate upward to the surface when they are released (if they were more dense than water, they would sink more).

We will treat the densities as relative to water, so it becomes rho_w=1.

If we distribute the product, and know that the density of B is higher than the density of A, and both are higher than the product of the densities, we have:

$\rho_w=1\\\\\frac{a_a}{a_b}=\frac{(1-\rho_a)}{(1-\rho_b)} \frac{\rho_b}{\rho_a}=\frac{\rho_b-\rho_a\rho_b}{\rho_a-\rho_a\rho_b}\\\\\\\rho_b>\rho_a>\rho_a\rho_b>0\\\\\\\frac{a_a}{a_b}=\frac{\rho_b-\rho_a\rho_b}{\rho_a-\rho_a\rho_b}>1\\\\a_a>a_b$

The acceleration of A is higher than the acceleration of B.

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4 years ago

Is the tropic of cancer north or south of the equator?

Daniel [21]

It’s North of the equator

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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:

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

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

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

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

7 0

3 years ago