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k0ka [10]
3 years ago
7

You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle

between A and E. (You will need a protractor to mcasure angles.) 1. Place the loop over the nails so that the number (# of field lines through surface) of field lines through it is a maximum, Determine the angle between A and E. Record both that angle and the number of field lines that pass through the loop. 2. Rotate the loop until there is one fewer row of nails passing through it. Determine the angle between A and E and record your measurement. Continue in this way until θ= 180. 3. On graph paper, plot a graph of n versus o (Let the number of field lines through the surface be a negative number for angles between 90 and 180.) When E and A were parallel, we called the quantity EA the electric flux through the surface. For the parallel case, we found that EA is proportional to the number of field lines through the surface. E. By what trigonometric function of θ must you multiply EA so that the product is proportional to the number of ficld lines through the area for any orientation of the surface? Rewrite the quantity described above as a product of just the vectors E and A.

Physics
1 answer:
nikitadnepr [17]3 years ago
4 0

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If  θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for  90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function  

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A  

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Gemiola [76]

Answer:

Time taken, T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

T\ cos\theta-mg=0

T=\dfrac{mg}{cos\theta}

Sum of forces in x direction,

T\ sin\theta=\dfrac{mv^2}{r}

mg\ tan\theta=\dfrac{mv^2}{r}.............(1)

Also, r=l\ sin\theta

Equation (1) becomes :

mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}

v=\sqrt{gl\ tan\theta.sin\theta}...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

T=\dfrac{2\pi r}{v}

Put the value of T from equation (2) to the above expression:

T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}

T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}

On solving above equation :

T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Hence, this is the required solution.

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3 years ago
You launch a ball at an angle of 35 degrees above the horizontal with an initial velocity of 38 m/s. What is the time the ball w
gayaneshka [121]

Vf=Vi+at

0=38+(-9.8)(?)

?=38-0+(-9.8)

?=28.2 s

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You exert a force of 25 newtons while you move a rock 15 meters. How much work did you perform?
leonid [27]
Work Done = Force x Distance Moved
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An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate
riadik2000 [5.3K]

Answer: 6.408(10)^{-19} C

Explanation:

This problem can be solved by the following equation:

\Delta K=q V

Where:

\Delta K=7.37(10)^{-17} J is the change in kinetic energy

V=115 V is the electric potential difference

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Finding q:

q=\frac{\Delta K}{V}

q=\frac{7.37(10)^{-17} J}{115 V}

Finally:

q=6.408(10)^{-19} C

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3 years ago
Which has the larger kinetic energy, a 10 g bullet fired at 400 m/s or a 80 kg bowling ball rolled at 6.5 m/s ?
mixer [17]
Formulae for Kinetic energy is:
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For comparison we need to have same units,thus we convert 10g into Kg.
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Which means that th Energy of the ball is more than the bullet.
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