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A 20 Ohm resistance is connected

frutty [35]

Answer: 225 V

Explanation:

Given:

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}$$$\therefore \text { Turns ratio }=\frac{N_{1}}{N_{2}}=\frac{7.5}{5}=\frac{3}{2}=3: 2$$$

$\text{Again taking,}\frac{N_{1}}{N_{2}}=\frac{V_{2}}{V_{1}}$ \\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, {$V_{1}=225 \mathrm{~V}$}}$

8 0

2 years ago

What kind of degree does an ultrasound tech. need?

Anna11 [10]

Educational Requirements. Training to become an ultrasound technician can be done through a formal education program or through military training. The most common training is an associate's degree program, although there are bachelor's degree and 1-year certificate programs available.

7 0

3 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.16

ycow [4]

We assign the variables: T as tension and x the angle of the string
The centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.

(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. 
(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. 
(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. 
(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.

This minimum v is obtained when T=0 and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.

4 0

3 years ago

Read 2 more answers

a load of 400 Newton is lifted by a first class lever in which the load is at the distance of 20 cm and the effort is at the dis

Len [333]

Answer:

solution,
Given
load =400N
ld=0.2m
ed=0.6m
effort =150N

Explanation:

efficiency =output work/input work ×100%

l×ld/e×ed×100%

400×0.2/150×0.6×100%

80/90×100%

88.89%ans

7 0

3 years ago

During the spin cycle of a washing machine, the clothes stick to the outer wall of the barrel as it spins at a rate as high as 1

Darya [45]

To answer the two questions, we need to know two important equations involving centripetal movement:

v = ωr (ω represents angular velocity in radians)

a = $\frac{v^{2}}{r}$

Let's apply the first equation to question a:

v = ωr

v = ((1800*2π) / 60) * 0.26

Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.

Back to the equation:

v = ((1800*2π)/60) * 0.26

v = (1800*2(3.14)/60) * 0.26

v = (11304/60) * 0.26

v = 188.4 * 0.26

v = 48.984

v = 49 (m/s)

Now that we know the linear velocity, we can find the centripetal acceleration:

a = $\frac{v^{2}}{r}$

a = $\frac{49^{2}}{0.26}$

a = 9234.6 (m/ $s^{2}$ )

Wow! That's fast!

We now have our answers for a and b:

a. 49 (m/s)

b. 9.2 * $10^{3}$ (m/ $s^{2}$ )

If you have any questions on how I got to these answers, just ask!

- breezyツ

5 0

3 years ago