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3 years ago

Why do you think a “reliable, scholarly” source is always stressed in academic writing?

Physics

1 answer:

Anvisha [2.4K]3 years ago

4 0

Answer:

PLS MARK ME AS BRAINLIEST

Explanation:

The need for reference of reliable scholarly source in academic writing (i.e. project) is to provide guide needed to confirm the writing of the scholar and to ascertain the writing are clear of plagiarism.

In academic writing, writers are required to state out the list of sources of information because:

1)To confirm authenticity of the writing

2)To prevent future plagiarism claim from the scholar.

3)To provide guide on how to trace the scholar writing

4)To complete the requirement of good academics writing

In conclusion, providing reliable scholarly source is the backbone of writing a perfect academics article, journal inclusive.

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Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago

With a bit of algebraic reasoning find your gravitational acceleration toward any planet of mass M a distance d from its center.

grandymaker [24]

The acceleration due to gravity is given as:

g = GM/r²

<h3>Derivation of gravitational acceleration:</h3>

According to Newton's second law of motion,

F = ma

where,

F = force

m = mass

a = acceleration

According to Newton's law of gravity,

F<em>g </em>= GMm/(r + h)²

F<em>g = </em>gravitational force

From Newton's second law of motion,

F<em>g </em>= ma

a = F<em>g</em>/m

We can refer to "a" as "g"

a = g = GMm/(m)(r + h)²

g = GM/(r + h)²

When the object is on or close to the surface, the value of g is constant and height has no considerable impact. Hence, it can be written as,

g = GM/r²

Learn more about gravitational acceleration here:

brainly.com/question/2142879

#SPJ4

5 0

2 years ago

The name used to describe the range of EM waves

Nimfa-mama [501]

Answer: the electromagnetic spectrum

Explanation:

I’m not sure but I think this is it!

8 0

3 years ago

The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.

klemol [59]

The resistance is 27.5 ohms

5 0

4 years ago

Read 2 more answers

If the resultant of two velocity vectors of equal magnitude is also of the same magnitude, then which statement must be correct?

Tamiku [17]

The correct option is C) The angle between the vectors is 120°.

Why?

We can solve the problem and find the correct option using the Law of Cosine.

Let A and B, the given two sides and R the resultant (sum),

Then,

$R=A=B$

So, using the law of cosines, we have:

$R^{2}=A^{2}+B^{2}+2ABCos(\alpha)\\ \\A^{2}=A^{2}+A^{2}+2*A*A*Cos(\alpha)\\\\0=A^{2}+2*A^{2}*Cos(\alpha)\\\\Cos(\alpha)=-\frac{A^{2}}{2*A^{2}}=-\frac{1}{2}\\\\\alpha =Cos(-\frac{1}{2})^{-1}=120\°$

Hence, we have that the angle between the vectors is 120°. The correct option is C) The angle between the vectors is 120°

Have a nice day!

4 0

3 years ago