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3 years ago

A circular wire loop of radius 15.0 cm carries a current of 2.60

1 answer:

5 0

Part (a): Magnetic dipole moment

Magnetic dipole moment = IA, I = Current, A = Area of the loop
Then,
Magnetic dipole moment = 2.6*π*0.15^2 = 0.184 Am^2

Part (b): Torque acting on the loop
T = IAB SinФ, where B = Magnetic field, Ф = Angle
Then,
T = Magnetic dipole moment*B*SinФ = 0.184*12*Sin 41 = 1.447 Nm

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Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

<u>For aluminium:</u>

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

<u>For copper:</u>

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

<u>Aluminium wire:</u>

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

<u>Copper wire:</u>

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

3 0

3 years ago

Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th

melomori [17]

The units of G must be C. m³ / ( kg s² )

<h3>Further explanation</h3>

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

$\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }$

<em>F = Gravitational Force ( Newton )</em>

<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>

<em>m = Object's Mass ( kg )</em>

<em>R = Distance Between Objects ( m )</em>

Let us now tackle the problem !

To find unit of Gravitational Constant can be carried out in the following way:

$F = G \frac{m_1 ~ m_2}{R^2}$

${[N]}= G\frac{{[kg]}{[kg]}}{{[m^2]}}$

${[kg ~ m / s^2]}= G \frac{{[kg^2]}}{{[m^2]}}$

$G = \frac{{[kg ~ m / s^2]}{[m^2]}} {{[kg^2]} }$

$G = \frac{{[kg ~ m^3 / s^2]}} {{[kg^2]} }$

$G = \frac{{[m^3 / s^2]}} {{[kg]} }$

$\boxed {G = \frac{{[m^3]}} {{[kg ~ s^2]} }}$

The unit of G must be $\large {\boxed {\frac{m^3} {kg ~ s^2 }}}$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

5 0

3 years ago

Read 2 more answers

mojhsa [17]

B. the distance the star is from Earth

Explanation:

The apparent magnitude of star is a function of its distance from the earth. It is one of the physical properties that is used to study a star.

The apparent magnitude of a star or other astronomical bodies is a measure of their brightness as seen from a location on the earth.

The apparent magnitude of a star depends on:

Distance of the star from the location on earth.
luminosity of the star
the particles along the part of the star and earth that cuts off the light the earth receives.

learn more:

Star luminosity brainly.com/question/9084808

#learnwithBrainly

5 0

3 years ago

What is the minimum energy required to lift an object weighing 200 Newtons at a height of 20 meters?

lakkis [162]

Work = force x distance

200 Newtons x 20 meters

= 4,000 Joules

4 0

3 years ago

What would happened if there is no invented of machine?

yuradex [85]

if there were no invention of machines then life would have been more difficult and simple works could be hard to do. Even now we are using our phones, sitting in a AC room interacting to eachother from different places. without the invention of machines simple things like transportation would have been difficult. There would be horses and donkey for the transportation. There would be no electricity,no internet, no transportation, not even c computers or mobile etc. The market for business will be smaller, the knowledge and news about world would be less.

so the problem would have been bigger than we can imagine. But one thing is that nature could survive lot more compared to what we have done till now by destroying nature.

4 0

1 year ago