Answer:
THE MASS OF THE LIQUID IS 22.5 g
Explanation:
Density = 0.180 g/cm3
Side length = 5 cm
Mass = unknown
To calculate the mass of the liquid, we use the formula:
Mass = density * volume
Volume of a cube or cuboid container = l^3
Volume = 5 ^3 = 125 cm3
So therefore, the mass of the liquid is:
Mass = 0.180 * 125
Mass = 22.5 g
In conclusion, the mass of the liquid in the container is 22.5 g
Answer: 225 V
Explanation:
<u>Given:</u>
Secondary voltage,V2 = 150v
Resistance is connected across secondary winding,∴R2 = 20Ω
Supply current, ie, I1 = 5A
![\text{Now, secondary current,} \ $I_{2}=\frac{V_{2}}{R_{2}}=\frac{150}{20}=7.5$$$\therefore I_{2}=7.5 \mathrm{~A}$$For Transformers, we have $\frac{N_{2}}{N_{1}}=\frac{V_{2}}{V_{1}}=\frac{I_{1}}{I_{2}}$ so, Taking $\frac{N_{2}}{N_{1}}=\frac{l_{1}}{l_{2}}$ inserting all given \& obtained values, we get $\frac{N_{2}}{N_{1}}=\frac{5}{7.5}$$$\therefore \text { Turns ratio }=\frac{N_{1}}{N_{2}}=\frac{7.5}{5}=\frac{3}{2}=3: 2$$](https://tex.z-dn.net/?f=%5Ctext%7BNow%2C%20secondary%20current%2C%7D%20%5C%20%24I_%7B2%7D%3D%5Cfrac%7BV_%7B2%7D%7D%7BR_%7B2%7D%7D%3D%5Cfrac%7B150%7D%7B20%7D%3D7.5%24%24%24%5Ctherefore%20I_%7B2%7D%3D7.5%20%5Cmathrm%7B~A%7D%24%24For%20Transformers%2C%20we%20have%20%24%5Cfrac%7BN_%7B2%7D%7D%7BN_%7B1%7D%7D%3D%5Cfrac%7BV_%7B2%7D%7D%7BV_%7B1%7D%7D%3D%5Cfrac%7BI_%7B1%7D%7D%7BI_%7B2%7D%7D%24%20so%2C%20Taking%20%24%5Cfrac%7BN_%7B2%7D%7D%7BN_%7B1%7D%7D%3D%5Cfrac%7Bl_%7B1%7D%7D%7Bl_%7B2%7D%7D%24%20inserting%20all%20given%20%5C%26%20obtained%20values%2C%20we%20get%20%24%5Cfrac%7BN_%7B2%7D%7D%7BN_%7B1%7D%7D%3D%5Cfrac%7B5%7D%7B7.5%7D%24%24%24%5Ctherefore%20%5Ctext%20%7B%20Turns%20ratio%20%7D%3D%5Cfrac%7BN_%7B1%7D%7D%7BN_%7B2%7D%7D%3D%5Cfrac%7B7.5%7D%7B5%7D%3D%5Cfrac%7B3%7D%7B2%7D%3D3%3A%202%24%24)
![\text{Again taking,}\frac{N_{1}}{N_{2}}=\frac{V_{2}}{V_{1}}$ \\So, \\$V_{1}=\frac{N_{1}}{N_{2}} V_{2}\\$ inserting all values, we get, $V_{1}=\frac{7.5}{5} * 150=225$\\Therefore, \fbox{Primary voltage, {$V_{1}=225 \mathrm{~V}$}}](https://tex.z-dn.net/?f=%5Ctext%7BAgain%20taking%2C%7D%5Cfrac%7BN_%7B1%7D%7D%7BN_%7B2%7D%7D%3D%5Cfrac%7BV_%7B2%7D%7D%7BV_%7B1%7D%7D%24%20%5C%5CSo%2C%20%5C%5C%24V_%7B1%7D%3D%5Cfrac%7BN_%7B1%7D%7D%7BN_%7B2%7D%7D%20V_%7B2%7D%5C%5C%24%20inserting%20all%20values%2C%20we%20get%2C%20%24V_%7B1%7D%3D%5Cfrac%7B7.5%7D%7B5%7D%20%2A%20150%3D225%24%5C%5CTherefore%2C%20%5Cfbox%7BPrimary%20voltage%2C%20%7B%24V_%7B1%7D%3D225%20%5Cmathrm%7B~V%7D%24%7D%7D)
Answer:
the answer is the first letter in the alphabet