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madreJ [45]
3 years ago
14

For a moon orbiting its planet, rp is the shortest distance between the moon and its planet andra is the longest distance betwee

n the moon and its planet. What is a moon's orbital eccentricity if rp is equal to 0.27ra?
A. 0.65

B. 0.27

C. 0.48

D. 0.57

D. IS THE CORRECT ANSWER!
Physics
1 answer:
Natasha2012 [34]3 years ago
3 0

Answer: D. 0.57

Explanation:

The formula to calculate the eccentricity e of an ellipse is (assuming the moon's orbit in the shape of an ellipse):

e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}}

Where:

r_{a} is the apoapsis (the longest distance between the moon and its planet)

r_{p}=0.27 r_{a} is the periapsis (the shortest distance between the moon and its planet)

Then:

e=\frac{r_{a}-0.27 r_{a}}{r_{a}+0.27 r_{a}}

e=\frac{0.73 r_{a}}{1.27 r_{a}}

e=0.57 This is the moon's orbital eccentricity

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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

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3 years ago
An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase
goldenfox [79]

Answer:

changing the magnetic field more rapidly

Explanation:

According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.

Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.

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Explanation:

Hope this helps,

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As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road i
Darya [45]
<span>As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road is kinetic friction.

We exert force to move the object from rest and in this case, static friction works. But, when the object comes in motion, then kinetic friction works. Here, since the car is driving without slipping means, kinetic friction acts on it. Its also called sliding or dynamic friction.</span>
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Answer:

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