Please explain how to solve it!!!

Atomic nuclei of almost all elements consist of

wolverine [178]

<h2>Answer: protons and neutrons. </h2>

The atomic nuclei of almost all elements consist of protons and neutrons.

The nucleus of an atom has very small dimensions. However, it <u>occupies its central part and concentrates more than 99% of its total mass. </u>

It is in the nucleus that the protons (positive charge) and neutrons (neutral charge) are found.

4 0

4 years ago

S Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to

denpristay [2]

The diameter of the wire is 2.8 * 10^-3 m.

<h3>What is the length?</h3>

Mass of the wire = 1.0 g or 1 * 10^-3 Kg

Resistance = 0.5 ohm

Resistivity of copper = 1.7 * 10^-8 ohm meter

Density of copper = 8.92 * 10^3 Kg/m^3

V = m/d

But v = Al

Al = m/d

A = m/ld

Resistance = ρl/A

= ρl/m/ld =

l^2 = Rm/ρd

l = √ Rm/ρd

l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3

l = 1.82 m

A = πr^2

Also;

A = m/ld

A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3

A = 6.2 * 10^-5 m^2

r^2 = A/ π

r = √A/ π

r = √6.2 * 10^-5 m^2/3.142

r = 1.4 * 10^-3 m

Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m

Learn more about resistivity:brainly.com/question/14547003

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Missing parts;

Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?

5 0

2 years ago

Which of the following is an example of a dry lubricant? oil grease graphite petroleum

Naddika [18.5K]

Answer:

Dry lubricants or solid lubricants are materials that, despite being in the solid phase, are able to reduce friction between two surfaces sliding against each other without the need for a liquid oil medium.

Explanation:

7 0

3 years ago

Read 2 more answers

A scientist testing to see how a cat “always” lands on their feet drops a cat upside down out of a third story window at a heigh

makkiz [27]

<u>Answer: </u>

Cat has 2.02 seconds to right itself.

<u>Explanation: </u>

Initial height of cat from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of cat in vertical direction = 0 m/s, acceleration = acceleration due to gravity = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So, cat has 2.02 seconds to right itself.

7 0

3 years ago

Find the value of currents through each branch

Irina-Kira [14]

Answer:

the branch currents are as follows:

top left: I2 = 0.625 A

middle left: I1 = 2.500 A

bottom left: I1-I2 = 1.875 A

top center: I2+I3 = 2.500 A

bottom center: I2+I3-I1 = 0 A

right: I3 = 1.875 A

Explanation:

You can write the KVL equations:

Top left loop:

I2(4) +(I2 +I3)(2) +I1(1) = 10

Bottom left loop:

(I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10

Right loop:

(I2+I3)(2) +(I2+I3-I1)(2) = 5

In matrix form, the equations are ...

$\left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]$

These equations have the solution ...

$\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]$

This means the branch currents are as follows:

top left: I2 = 0.625 A

middle left: I1 = 2.500 A

bottom left: I1-I2 = 1.875 A

top center: I2+I3 = 2.500 A

bottom center: I2+I3-I1 = 0 A

right: I3 = 1.875 A

_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

(I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

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It helps to be familiar with the formulas for resistors in series and parallel.

8 0

3 years ago