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dexar [7]
3 years ago
5

An airplane is initially flying horizontally (not gaining or losing altitude), and heading exactly North. Suppose that the earth

's magnetic field at this point is also exactly horizontal, and points from South to North (it really does!). The airplane now starts a different motion. It maintains the same speed, but gains altitude at a constant rate, still heading North. The magnitude of the electric potential difference between the wingtips has... [Remember that the airplane is made of metal, and is a conductor] Group of answer choices
Physics
1 answer:
yanalaym [24]3 years ago
3 0

Note: The answer choices are :

a) Increased

b) Decreased

c) stayed the same

Answer:

The correct option is Increased

The magnitude of the electric field potential difference between the wingtips increases.

Explanation:

The magnitude of the electric potential difference is the induced emf and is given by the equation:

emf = l (v \times B)

where l = length

v = velocity

B = magnetic field

As the altitude of the airplane increases, the magnetic flux becomes stronger, the speed of the airplane becomes perpendicular to the magnetic field, i.e. v \times B = vB sin90 = vB\\ ,

the induced emf = vlB, and thus increases.

The magnitude of the electric field potential difference between the wingtips increases

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When an object is placed 110 cm from a diverging thin lens, its image is found to be 55 cm from the lens. The lens is removed, a
Kazeer [188]

Explanation:

Given that,

Object distance u= -110 cm

Image distance v= 55 cm

We need to calculate the focal length for diverging lens

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{-f}=\dfrac{1}{55}-\dfrac{1}{-110}

\dfrac{1}{f}=-\dfrac{3}{110}

f=-36.6\ cm

The focal length of the diverging lens is 36.6 cm.

Now given a thin lens with same magnitude of focal length 36.6 cm is replaced.    

Here, The object distance is again the same.

We need to calculate the image distance for converging lens

Using formula of lens

\dfrac{1}{36.6}=\dfrac{1}{v}-\dfrac{1}{-110}

Here, focal length is positive for converging lens

\dfrac{1}{v}=\dfrac{1}{36.6}-\dfrac{1}{110}

\dfrac{1}{v}=\dfrac{367}{20130}

v=54.85\ cm

The distance of the image is 54.85 cm from converging lens.

Hence, This is the required solution.

5 0
3 years ago
you push a ball to star it rolling along a "perfectly frictionless" surface. How far will the ball roll
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A ball has a diameter of 3.79 cm and average density of 0.0838 g/cm3.
suter [353]

Answer: 0.258 N

Explanation:

As the density of the object is much less than the density of water, it’s clear that the buoyant force, is greater than the weight of the object, which means that in normal conditions, it would float in water.

So, in order to get the ball submerged in water, we need to add a downward force, that add to the weight, in order to compensate the buoyant force, as follows:

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Fb= δH20* 4/3*π*(d/2)³  * g

Fg = δb* 4/3*π*(d/2)³ *g

F= (δH20- δb) * 4/3*π*(d/2)³*g

Replacing by the values of the densities, and the ball diameter, we finally get:

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