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2 years ago

Why are the spheres representing nitrogen and oxygen different colors

Physics

1 answer:

erastovalidia [21]2 years ago

6 0

Spheres representing nitrogen and oxyfgen are different colors so you can easily tell them apart.

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Two objects that are not initially in thermal equilibrium are placed in close contact. After a while, the temperature of the cod

Dima020 [189]

Answer:

If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops, then the specific heats of both objects will be equal.

Explanation:

If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops when the two objects of same mass are brought into contact, then their specific heat capacity is equal.

We can prove this by the equation of heat for the two bodies:

According to given condition,

$\Delta T_1=\Delta T_2$

$\frac{Q_1}{m_1.c_1} = \frac{Q_2}{m_2.c_2}$

when there is no heat loss from the system of two bodies then $Q_1=Q_2$

$\frac{1}{m.c_1} =\frac{1}{m.c_2}$

$\Rightarrow c_1=c_2$

Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.

The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:

when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.

the mass of colder object is half the mass of the hotter object while their specific heat is same.

3 0

2 years ago

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Sever21 [200]

Answer: $2.89\approx 2.9^{\circ}C/s$

Explanation:

Given

$Power\left ( P\right )=150 MW$

mass of core $\left ( m\right )=1.60\times 10^5 kg$

Average specific heat $\left ( C\right )=0.3349 KJ/kg^{\circ}C$

And rate of increase of temperature = $\frac{\mathrm{d}T}{\mathrm{d} t}$

Now

P= $mc\frac{\mathrm{d}T}{\mathrm{d} t}$

$150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}$

Thus $\frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}$

$\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s$

6 0

3 years ago

I need helpppppppppppppppppp!

allsm [11]

Answer:

i think its B

Explanation:

but check it before do it

3 0

3 years ago

Read 2 more answers

you throw an object straight up at 15m/s. how fast is it going when it gets back to the height from which you threw it?

laiz [17]

At the same speed because it will slow down as it approaches the peak then speed up as it goes down again

it will be going 15m/s when it gets to the same height if we neglect air resistance and the object doesn't hit something

8 0

2 years ago

As part

umka21 [38]

Answer:

Part a)

$a = 3.68 m/s^2$