Answer:
The average speed of the cyclist is 17.14 km/hr
Explanation:
Given;
total distance traveled by the cyclist's, d = 60 km
time taken by the cyclist, t = 3.5 hours
The average speed of the cyclist is given by;
average speed = total distance traveled / total time taken
average speed = 60 km / 3.5 hr
average speed = 17.14 km/hr
Therefore, the average speed of the cyclist is 17.14 km/hr
The question is: You have 500g of ethyl alcohol at a temperature of -40 ° C. How much heat is needed to transform it into steam at a temperature of 150ºC?
Answer: 233700 J heat is needed to transform ethyl alcohol into steam at a temperature of
to
.
Explanation:
Given: Mass = 500 g
Initial temperature = ![-40^{o}C](https://tex.z-dn.net/?f=-40%5E%7Bo%7DC)
Final temperature = ![150^{o}C](https://tex.z-dn.net/?f=150%5E%7Bo%7DC)
The standard value of specific heat of ethyl alcohol is
.
Formula used to calculate the heat energy is as follows.
![q = m \times C \times (T_{2} - T_{1})](https://tex.z-dn.net/?f=q%20%3D%20m%20%5Ctimes%20C%20%5Ctimes%20%28T_%7B2%7D%20-%20T_%7B1%7D%29)
where,
q = heat energy
m = mass of substance
C = specific heat
= initial temperature
= final temperature
Substitute the values into above formula as follows.
![q = m \times C \times (T_{2} - T_{1})\\= 500 g \times 2.46 J/g^{o}C \times [150 - (-40)]^{o}C\\= 233700 J](https://tex.z-dn.net/?f=q%20%3D%20m%20%5Ctimes%20C%20%5Ctimes%20%28T_%7B2%7D%20-%20T_%7B1%7D%29%5C%5C%3D%20500%20g%20%5Ctimes%202.46%20J%2Fg%5E%7Bo%7DC%20%5Ctimes%20%5B150%20-%20%28-40%29%5D%5E%7Bo%7DC%5C%5C%3D%20233700%20J)
Thus, we can conclude that 233700 J heat is needed to transform ethyl alcohol into steam at a temperature of
to
.
Answer:
<h2>Case i) if
![\omega L > \frac{1}{\omega c}](https://tex.z-dn.net/?f=%5Comega%20L%20%3E%20%5Cfrac%7B1%7D%7B%5Comega%20c%7D)
</h2><h2>So initially if the circuit is inductive in nature then its net impedance will decrease after this</h2><h2>Case ii) if
![\omega L < \frac{1}{\omega c}](https://tex.z-dn.net/?f=%5Comega%20L%20%3C%20%5Cfrac%7B1%7D%7B%5Comega%20c%7D)
</h2><h2>So initially if the circuit is capacitive in nature then its net impedance will increase after this</h2>
Explanation:
As we know that the impedance of the circuit is given as
![z = \sqrt{(\omega L - \frac{1}{\omega c})^2 + R^2}](https://tex.z-dn.net/?f=z%20%3D%20%5Csqrt%7B%28%5Comega%20L%20-%20%5Cfrac%7B1%7D%7B%5Comega%20c%7D%29%5E2%20%2B%20R%5E2%7D)
when we join another identical capacitor in parallel with previous capacitor in the circuit then we will have for parallel combination
![c_{eq} = c_1 + c_2](https://tex.z-dn.net/?f=c_%7Beq%7D%20%3D%20c_1%20%2B%20c_2)
so it is
![c_{eq} = 2c](https://tex.z-dn.net/?f=c_%7Beq%7D%20%3D%202c)
now we have
![z = \sqrt{(\omega L - \frac{1}{2\omega c})^2 + R^2}](https://tex.z-dn.net/?f=z%20%3D%20%5Csqrt%7B%28%5Comega%20L%20-%20%5Cfrac%7B1%7D%7B2%5Comega%20c%7D%29%5E2%20%2B%20R%5E2%7D)
Case i) if ![\omega L > \frac{1}{\omega c}](https://tex.z-dn.net/?f=%5Comega%20L%20%3E%20%5Cfrac%7B1%7D%7B%5Comega%20c%7D)
So initially if the circuit is inductive in nature then its net impedance will decrease after this
Case ii) if ![\omega L < \frac{1}{\omega c}](https://tex.z-dn.net/?f=%5Comega%20L%20%3C%20%5Cfrac%7B1%7D%7B%5Comega%20c%7D)
So initially if the circuit is capacitive in nature then its net impedance will increase after this
Then the force will also be doubled
Fair enough, but you'll have to tell us the volume of the bar first.