The resistance of the thermometer at room temperature is 15.04 ohms.
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<h3>What is a resistance thermometer?</h3>
A resistance thermometer is a type of thermometer that measures temperature through a change in resistance.
To calculate the resistance of the thermometer at room temperature, we use the formula below.
Formula:
- 100/27 = 2/(x-14.5)..............Eqquation 1
Where:
- x = Resistance of the thermometer at room temperature
Make x the subject of the equation
- x = [(27×2)/100]+14.5
- x = (54/100)+14.5
- x = 0.54+14.5
- x = 15.04 ohms.
Hence, The resistance of the thermometer at room temperature is 15.04 ohms.
Learn more about thermometers here: brainly.com/question/1531442
Emf = d (phi-B) / dt
<span>B dA/dt, where dA/dt is the area swept out by the wire per unit time. </span>
<span>0.88 V = (0.075 N/(A m)) (L)(4.20 m/s), so </span>
<span>L = (0.88 J/C) / [ (0.075 N s/C m)(4.2 m/s) ] = about 3 meters</span>
Answer:
Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc) items 3 and 4 are eliminated.
Again item 2 refers to bonds between atoms and is eliminated.
This leaves only item 1.
Nuclear forces are very short range forces between components of the nucleus.
Weak nuclear forces are trillions of times smaller than strong forces.
Gravitational forces are much much smaller than the weak nuclear force.
The discovery which Carnot made was that THE DIFFERENCE IN THE TEMPERATURES BETWEEN THE HOT AND THE COLD RESERVOIRS DETERMINE HOW WELL A HEAT ENGINE WOULD WORK.
Sadi Carnot was a French engineer, He proposed a theoretical thermodynamic cycle in 1824. In his cycle, Said hold that the efficiency of a heat engine depends on the temperature difference between its hot reservoir and cold reservoir.
Answer:
4.6 m
Explanation:
First of all, we can find the frequency of the wave in the string with the formula:
where we have
L = 2.00 m is the length of the string
T = 160.00 N is the tension
is the mass linear density
Solving the equation,
The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.
The n=1 mode (fundamental frequency) of an open-open tube is given by
where
v = 343 m/s is the speed of sound
Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:
