All categories

3 years ago

A circuit in which two or more paths are connected to the voltage source

1 answer:

7 0

That is a parallel circuit and unlike a series circuit it has multiple paths so if a light bulb goes out it can still flow back to the power source. <span />

You might be interested in

What dose nuclear reactions produce?

aleksandrvk [35]

Answer:

a new chromebook for you and you will get to know the other one that

7 0

2 years ago

A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di

BlackZzzverrR [31]

Answer:

$r =3 *\sqrt{2} = 4.24 cm$

Explanation:

given data

Radius of sphere 3.0 cm

charge Q = 2.0 m C

We know that maximum electric field is given as

$E_{MAX}= \frac{KQ}{r^{2}}$

electric field inside the sphere can be determine by using below relation

$\frac{KQ}{r^{2}}= \frac{1}{2}*\frac{KQ}{R^{2}}$

$r = \sqrt{2}R$

$r =3 *\sqrt{2} = 4.24 cm$

4 0

3 years ago

PLEASE HELP ME WITH MY HOMEWORK!!!!!

IgorC [24]

Known :

l = 7 cm

w = 4 cm

Asked :

h = ...?

Answer :

V = B triangle × h (long)

35 = ½ × 4 × h × 7

35 = ½ × 28 × h

35 = 14 h

h = 35 ÷ 14

h = 2,5 cm

Sorry if I am wrong, I only study

5 0

3 years ago

Explian the construction of electric motor?

Nataly_w [17]

Answer:

Hope the above picture might help you :)

7 0

3 years ago

A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w

barxatty [35]

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

$Moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}$

$Moles\ of\ AgBr= 0.1225\ mol$

The reaction taking place is:

$MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4$

From the reaction,

4 moles of AgBr is produced when 1 mole of $MBr_4$ undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of $MBr_4$ undergoes reaction

0.1225 mole of AgBr is produced when $\frac {1}{4}\times 0.1225$ mole of $MBr_4$ undergoes reaction

Moles of $MBr_4$ got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

So, Molar mass of $MBr_4$ = x + 4 × 79.904 g/mol = 319.616 + x g/mol

Thus,

$0.030625 = \frac{12.5843}{319.616 + x}$

Solve for x,

we get, x = 91.2999 g/mol

<u>The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.</u>

8 0

2 years ago