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3 years ago

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The air in living room has a mass of 72 kg and a specific heat of 1010 J/(kg°C). What is the approximate change in thermal energ

y of the air when it warms from 15°C to 20°C?

1 answer:

ICE Princess25 [194]3 years ago

3 0

Answer:

d

Explanation:

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An object weigh 40N in air ,weigh 20N when submerged in water,and 30N when submerged in a liquid of unknown liquid density.what

muminat

Answer:

The density is $\rho_u =500 kg /m^3$

Explanation:

From the question we are told that

The weight in air is $W_a = 40 \ N$

The weight in water is $W_w = 20 \ N$

The weight in a unknown liquid is $W_u = 30 \ N$

Now according to Archimedes principle the weight of the object in water is mathematically represented as

$W_w = W_a -m _w g$

Where $m_w$ is he mass of the water displaced

substituting value

$m_w g = 40 -20$

$m_w g = 20 \ N --- (1)$

Now according to Archimedes principle the weight of the object in unknown is mathematically represented as

$W_u = W_a -m _u g$

Where $m_u$ is he mass of the unknown liquid displaced

substituting value

$m_u g = 40 -30$

$m_u g = 10 \ N ---(2)$

dividing equation 2 by equation 1

$\frac{m_ug}{m_wg} = \frac{10}{20}$

$\frac{m_u}{m_w} = \frac{1}{2}$

=> $m_u = 0.5 m_w$

Now since the volume of water and liquid displaced are the same then

$\rho _u = 0.5 \rho_w$

This because

$density = \frac{mass}{volume}$

So if volume is constant

mass = constant * density

Where $\rho_u$ is the density of the liquid

and $\rho_ w$ is the density of water which is a constant with a value $\rho_w = 1000 kg/m^3$

So

$\rho_u = 1000*0.5$

$\rho_u =500 kg /m^3$

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3 years ago

Electronic brainstorming, also called _____, is a technique used to help members of a group come together over a computer networ

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Answer:

Brainwriting

Explanation:

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3 years ago

A mirror with a flat surface is a

USPshnik [31]

A- plane mirror because its surface is plane

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3 years ago

Read 2 more answers

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

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3 years ago

How do the chemical properties of the halogens compare to those of the noble gases?

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Halogens<span> are extremely reactive elements because they need one more electron to gain a full octet of valence electrons, whereas the </span>noble gases<span>are extremely unstable because they already have their full octet.</span>

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3 years ago