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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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Ber [7]

Answer: -6.80\ ms^{-2}

Explanation:

We know that the formula for acceleration is given by:

a=\dfrac{v-u}{t} , where v = Final velocity

u= Initial velocity

Given :  The driver of a car traveling 110 km/h slams on the brakes so that the car undergoes a constant acceleration.

i.e. u=   110 km/h =\dfrac{110\times1000}{3600}\approx35.6\ m/s  [∵  1 km= 100 meters and 1 hour = 3600 seconds]

v=  0  m/s ( At brake , final velocity becomes 0)

t=4.5 seconds  

Substitute all the values in the formula , we get

a=\dfrac{0-30.56}{4.5}\approx-6.80\ ms^{-2}

Hence, the average acceleration of the car during braking is -6.80\ ms^{-2}.

5 0
2 years ago
You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.220 kg. All of the ice melts,
g100num [7]

Answer:

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Explanation:

As we know that here final equilibrium temperature is 0 degree C

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so we have

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serg [7]
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joja [24]

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tekilochka [14]

By Newton's second law, the net force on the object is

∑ <em>F</em> = <em>m</em> <em>a</em>

∑ <em>F</em> = (2.00 kg) (8 <em>i</em> + 6 <em>j</em> ) m/s^2 = (16.0 <em>i</em> + 12.0 <em>j</em> ) N

Let <em>f</em> be the unknown force. Then

∑ <em>F</em> = (30.0 <em>i</em> + 16 <em>j</em> ) N + (-12.0 <em>i</em> + 8.0 <em>j</em> ) N + <em>f</em>

=>   <em>f</em> = (-2.0 <em>i</em> - 12.0 <em>j</em> ) N

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