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Gennadij [26K]
2 years ago
15

Can someone tell me a procedure to carry out the experiment that steam hotter than water

Physics
1 answer:
sergejj [24]2 years ago
4 0

Answer:

<u><em>STEAM IS BASICALLY A GAS PRODUCED OR A VAPOR, MORE ACCURATELY PRODUCED WHEN WATER HEATS UP TO MORE THAN 100 DEGREES CELSIUS. WHEN WATER BOLIS MORE THAN 100 DEGREES CELCIUS IT CAN NOT BE HEATED UP ANYMORE AND WILL BE RELEASED AS VAPOR, WHICH IS THE HOT VAPOUR PRODUCED THAT IS HOOTER THAN WATER. STEAM IS THE RESULT OF A HOT EXPERIMENT WITH WATER AFTER HEATING IT, THUS MAKING IT MORE HOTTER THAN BOILED WATER.</em></u>

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An earthquake sends a seismic wave through the ground. What is the<br> medium? ANSWER QUICK!
konstantin123 [22]

Answer:

THE GROUND IS THE MEDIUM OF SEISMIC WAVES

6 0
3 years ago
How far will you travel if you run for 10. minutes at 2.0 m/s?
oee [108]
We know that 1 minute= 60 seconds (or 1 min= 60 s).

10 min* (60 s/ 1 min)* (2.0 m/ 1 s)= 1,200 m.
(Note that the units cancel out so you get the answer)

The final answer is 1,200 m.

Hope this helps~
6 0
3 years ago
You are given a sample of metal and asked to determine its specific heat. You weight the sample and find that its weight is 28.4
lidiya [134]

Answer:

The samples specific heat is 14.8 J/kg.K

Explanation:

Given that,

Weight = 28.4 N

Suppose, heat energy E=1.25\times10^{4}\ J

Temperature = 18°C

We need to calculate the samples specific heat

Using formula of specific heat

Q=mc\Delta T

c=\dfrac{Q}{m\Delta T}

Where, m = mass

c = specific heat

\Delta T = temperature

Q = heat

Put the value into the formula

c=\dfrac{1.25\times10^{4}}{\dfrac{28.4}{9.8}\times(18+273)}

c=14.8\ J/kg. K

Hence, The samples specific heat is 14.8 J/kg.K

8 0
3 years ago
Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying
agasfer [191]

Answer:

The charge density in the system is 4.25*10^4C/m

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C

We need to asume here the number of free electrons in a copper conductor, at which is generally of 8.5 *10^{28}m^{-3}

The equation to find the current is

I = VenA

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)

I= 22.11a

Now to find the linear charge density, we know that

I = \frac{Q}{t} \rightarrow Q = It

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

V_d = \frac{l}{t} \rightarrow l = V_d t

The charge density is defined as

\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}

Replacing our values

\lambda = \frac{22.11}{5.20*10{-4}}

\lambda= 4.25*10^4C/m

Therefore the charge density in the system is 4.25*10^4C/m

5 0
4 years ago
If you see a rainbow near the time of sunset, where in the sky will the rainbow be?
Igoryamba
Closest to the sun f thats a choice or closest to the atmoshpere
8 0
4 years ago
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