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Nimfa-mama [501]
3 years ago
7

What is the low end of the range of surface temperature of white stars

Physics
1 answer:
Lelechka [254]3 years ago
8 0
The way astronomers determine the surface temperature<span> of a </span>star<span> which is determined by the spectral lines present in a </span>stars<span> spectrum</span>
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The stopcock connecting a 2.14 L bulb containing oxygen gas at a pressure of 8.19 atm, and a 9.84 L bulb containing krypton gas
marshall27 [118]

Answer : The final pressure of the system in atm is, 3.64 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1+P_2V_2=P_fV_f

where,

P_1 = first pressure = 8.19 atm

P_2 = second pressure = 2.65 atm

V_1 = first volume = 2.14 L

V_2 = second volume = 9.84 L

P_f = final pressure = ?

V_f = final volume = 2.14 L  + 9.84 L = 11.98 L

Now put all the given values in the above equation, we get:

8.19atm\times 2.14L+2.65atm\times 9.84L=P_f\times 11.98L

P_f=3.64atm

Therefore, the final pressure of the system in atm is, 3.64 atm

4 0
3 years ago
Some gamma ray bursts are hypothesized to come from mergers of neutron stars or black holes. if this hypothesis is correct, what
Umnica [9.8K]
We should see (and have now detected with LIGO) gravitational waves
6 0
3 years ago
When is the best time to take a resting heart break
Olegator [25]

Answer:

I believe D

Explanation:

You need to have a more accurate reading and you want to test it multiple times throughout the week though to get a base resting rate.

I hope this is correct good luck!

5 0
3 years ago
5)
miskamm [114]
B . I hope this is right
4 0
3 years ago
How much ice (at 0°C) must be added to 1.90 kg of water at 79 °C so as to end up with all liquid at 8 °C? (ci = 2000 J/(kg.°
muminat

m = mass of the ice added = ?

M = mass of water = 1.90 kg

c_{w} = specific heat of the water = 4186 J/(kg ⁰C)

c_{i}  = specific heat of the ice = 2000 J/(kg ⁰C)

L_{f} = latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg

T_{ii}  = initial temperature of ice = 0 ⁰C

T_{wi} = initial temperature of water = 79 ⁰C

T = final equilibrium temperature = 8 ⁰C

using conservation of heat

Heat gained by ice = Heat lost by water

m c_{w}  (T - T_{ii} ) + m L_{f}  = M c_{w}  (T_{wi} - T)

inserting the values

m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)

m = 1.53 kg

4 0
3 years ago
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