Ask question

Ask question

All categories

3 years ago

10

How would you draw a parallel circuit with two resistors and an ammeter so the ammeter measures total current?

1 answer:

DiKsa [7]3 years ago

7 0

It's really hard to explain in words.
See the attached diagram.

You might be interested in

today in my world geography class we were talking about police violence and my teacher asked me "who can watch the polices and w

Blababa [14]

Well, it almost sounds like a rhetorical question. Who can? Well, probably someone higher in power with the help of technology. They already have cameras being built into the suits due to the reports of unfair treatment from police officers. We could possibly have people from the government watch over footage from the cameras or have supervisors go with them. Anyway, that's my possible answer/opinion to that subject. Maybe that can give you some insight (-:

4 0

3 years ago

.What is geothermal heat? How does Iceland get its energy for heat and electricity?

Nataly [62]

Geothermal heat, is heat being pulled from the earth loop, creating warm air for us. Iceland uses geothermal heating for heat and electricity.

6 0

3 years ago

Where does Canada export it's surplus oil

Xelga [282]

They export from the United States.

8 0

3 years ago

Read 2 more answers

What are wandering or roving tribes called? The answer has to be six letters long.

Setler79 [48]

nomads
nomads
nomads
nomads

8 0

3 years ago

Read 2 more answers

What is the value of X6? <br><br> Show the solution.

wariber [46]

Answer : The value of $x_6$ is $\sqrt{7}$ .

Explanation :

As we are given 6 right angled triangle in the given figure.

First we have to calculate the value of $x_1$ .

Using Pythagoras theorem in triangle 1 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_1)^2=(1)^2+(1)^2$

$x_1=\sqrt{(1)^2+(1)^2}$

$x_1=\sqrt{2}$

Now we have to calculate the value of $x_2$ .

Using Pythagoras theorem in triangle 2 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_2)^2=(1)^2+(X_1)^2$

$(x_2)^2=(1)^2+(\sqrt{2})^2$

$x_2=\sqrt{(1)^2+(\sqrt{2})^2}$

$x_2=\sqrt{3}$

Now we have to calculate the value of $x_3$ .

Using Pythagoras theorem in triangle 3 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_3)^2=(1)^2+(X_2)^2$

$(x_3)^2=(1)^2+(\sqrt{3})^2$

$x_3=\sqrt{(1)^2+(\sqrt{3})^2}$

$x_3=\sqrt{4}$

Now we have to calculate the value of $x_4$ .

Using Pythagoras theorem in triangle 4 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_4)^2=(1)^2+(X_3)^2$

$(x_4)^2=(1)^2+(\sqrt{4})^2$

$x_4=\sqrt{(1)^2+(\sqrt{4})^2}$

$x_4=\sqrt{5}$

Now we have to calculate the value of $x_5$ .

Using Pythagoras theorem in triangle 5 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_5)^2=(1)^2+(X_4)^2$

$(x_5)^2=(1)^2+(\sqrt{5})^2$

$x_5=\sqrt{(1)^2+(\sqrt{5})^2}$

$x_5=\sqrt{6}$

Now we have to calculate the value of $x_6$ .

Using Pythagoras theorem in triangle 6 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_6)^2=(1)^2+(X_5)^2$

$(x_6)^2=(1)^2+(\sqrt{6})^2$

$x_6=\sqrt{(1)^2+(\sqrt{6})^2}$

$x_6=\sqrt{7}$

Therefore, the value of $x_6$ is $\sqrt{7}$ .

5 0

3 years ago