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OverLord2011 [107]
3 years ago
9

A less common type of radioactive particle emitted is the positron. If potassium-38 forms argon-38 by positron emission, the num

ber of a positron is and the charge is
Physics
1 answer:
aleksandr82 [10.1K]3 years ago
4 0
Potassium goes to argon so that a proton in the K nucleus "vanishes" into a positron (positive electron ?) but keeps the same atomic mass at 38. ? I'd say that the number of the positron was 0, and the charge was +1 electronic charge.
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Two resistors with values of 6.0 ohms and 12 ohms are connected in parallel. This combination is connected in series with a 2.0
ivanzaharov [21]

Answer:

4A

Explanation:

According to ohm's law;

E = IRt where;

E is the source voltage = 24volts

I is the total current flowing in the circuit = ?

Rt is the total effective resistance in the circuit.

To find Rt, we will resolve the resistors in parallel first.

Since 6ohms and 12ohms resistors are in parallel, their effective resistance will give;

1/R = 1/6+1/12

1/R= 2+1/12

1/R = 3/12

3R = 12

R = 4ohms.

This resistor will now be in series with the 2.0ohms resistor to finally have;

Rt = 4+2

Rt = 6ohms

From the ohms law formula;

I = E/Rt

I = 24/6

I = 4Amperes

The total current in the circuit is 4A

This same currents will flow in the 2ohms resistor since same current flows in a series connected resistors.

3 0
3 years ago
Read 2 more answers
Help<br><br><br> What are energy and work
mylen [45]

Answer:

<em>In physics, energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat, the object. Energy is a conserved quantity; the law of conservation of energy states that energy can be converted in form, but not created or destroyed.</em>

<em />

<em>In physics, energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat, the object. Energy is a conserved quantity; the law of conservation of energy states that energy can be converted in form, but not created or destroyed.</em>

Explanation:

<h2><u><em>HOPE THIS HELPS</em></u></h2>
8 0
3 years ago
If the moon were twice as far from years as it is now the following would be true
zubka84 [21]
The question is incomplete.

The distance between the Moon and Earth influences: 1) the attractive gravitational force between them, 2) the tides, 3) the eclipses, 4) the period of each full turn of the moon around the Earth.

Assuming the question refers to the gravitational attraction, we must use the fact that, as per, Newton's Universal Gravitaional Law, the attractive force between the two bodies is inversely related to the square distance that separates them.

Then, if the Moon were twice as far, the gravitational pull would be one fourth (1/4) of actual pull.

7 0
3 years ago
How would the self-inductance, l, of a coil change if you would increase its radius by a factor of two and increase its length b
pochemuha

The self-inductance of a coil will change by 8 times its original value by increasing its radius value by 2 and increasing the length of the coil by 2.

Self-Inductance: -

The definition of self-inductance is the induction of a voltage in a wire that carries current when the current in the wire is changing. In the instance of self-inductance, the circuit itself induces a voltage through the magnetic field produced by a changing current.

We know that the self-inductance of the coil is denoted by: -

L= µ *π*(r)^2*(N)^2*l

Where

L= Self-Inductance of the coil

µ= Magnetic Permeability Constant

r= Radius of the coil

l= Length of the coil

N= Number of turns of the coil

Here Self-inductance of the coil is directly proportional to the length of the coil and the square of the radius of the coil.

So,

On increasing the radius of the coil by a factor of 2 and the length of the coil by 2 the self-inductance of the coil increases by 8 times its original value.

Learn more about Self-Inductance here: -

" brainly.com/question/15293029 "

#SPJ4

3 0
2 years ago
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton
pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx  

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:


F = K * \Delta{x}

2\:N = 4\:N/cm*\Delta{x}

4\:N/cm*\Delta{x} = 2\:N

\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}

simplify by 2

\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}

\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark

Answer:

B.) 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

7 0
3 years ago
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