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marysya [2.9K]
3 years ago
7

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?
Physics
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:

5 0
3 years ago
A mirror with a flat surface is a
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A- plane mirror because its surface is plane
7 0
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.

7 0
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>
8 0
3 years ago
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