A smoke alarm beeps twice per minute. What is the period and frequency 1 po

Suppose that a rectangular toroid has 1,500 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

V125BC [204]

Answer:

Current in the toroid will be $I=17.32\times 10^{-3}A$

Explanation:

We have given number of winding in rectangular toroid N = 1500

Self inductance of toroid L = 0.06 H

Magnetic energy stored in toroid $E=9\times 10^{-16}J$

We have to find the current in the toroid

Magnetic energy stored is equal to $E=\frac{1}{2}Li^2$

$9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2$

$I=17.32\times 10^{-3}A$

So current in the toroid will be $I=17.32\times 10^{-3}A$

5 0

4 years ago

1+1=69 how is it possible

o-na [289]

Answer:

you add 67 to 1+1 is this clear also chicken nuggets

8 0

3 years ago

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The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material o

zavuch27 [327]

Answer:

pp

Explanation:

7 0

3 years ago

A bare helium nucleus has two positive charges and a mass of 6.64×10−27kg(a) Calculate its kinetic energy in joules at 2.00% of

grandymaker [24]

To solve this problem we will apply the concepts related to kinetic energy and energy conservation. The kinetic energy will be expressed in terms of mass and speed, as well as load and voltage. From this last expression we will find the charges by electron and by Helium nucleus.

a ) Kinetic Energy is given as

$KE = \frac{1}{2} mv^2$

Replacing with our values we have that

$KE = \frac{1}{2} (6.64*10^{-27})(2.0\% (3.00*10^8))^2$

$KE = 1.1935*10^{-13}J$

Therefore the kinetic energy of the helium nucleus is $1.1935*10^{-13}J$

PART B) Now for calculate the electron volts we use the kinetic energy as a expression between the charge and the voltage, that is

$KE = qV$

Here,

q = Charge of an electron

V = Voltage

Rearranging to find the potential we have,

$V = \frac{KE}{q}$

$V = \frac{1.19*10^{-13}}{1.6*10^{-19}}$

$V = 743750eV$

Therefore the kinetic energy in electron vols is $743750eV$

PART C) Applying the same relationship but now using the Helium core load, we will have to

$KE = QV$

Here,

Q = Charge of a helium nucleus

V = Voltage

Rearranging to find the potential we have

$V = \frac{KE}{Q}$

But we need to note that the charge is equal to the number of charge for the unit charge, then

$Q = \text{No. Charge} \times \text{Unit Charge}$

$Q = (2)(1.6*10^{-19}C)$

$Q = 3.2*10^{-19}C$

Now replacing we have that

$V= \frac{1.19*10^{-13}}{32*10^{-19}}$

$V = 371875V$

Therefore the voltage applied is 371875V

5 0

3 years ago

When studying atoms, scientists can ignore the ______ force between charged particles that make up the atoms because it is many

Elenna [48]

Answer:

When studying atoms, scientists can ignore the intermolecular electrostatic forces between charged particles that make up the atoms because it is many millions of times smaller than other forces in the atom. Gravitational Constant Normal Fricitional

3 0

3 years ago

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