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Zinaida [17]
3 years ago
11

Consider the uniform electric field \vec{E} =(4.00~\hat{j}+3.00~\hat{k})\times 10^3~\text{N/C} ​E ​⃗ ​​ =(4.00 ​j ​^ ​​ +3.00 ​k

​^ ​​ )×10 ​3 ​​ N/C. What is its electric flux through a circular area of radius 2.66 m that lies in the xy-plane?
Physics
1 answer:
Tema [17]3 years ago
6 0

Answer:

Electric flux, \phi=6.668\times 10^4\ Nm^2/C

Explanation:

It is given that,

Electric field, E=(4j+3k)\times 10^3\ N/C

We need to find the electric flux through a circular area of radius 2.66 m that lies in the xy-plane. A=Ak

The electric flux is given by :

\phi=E{\cdot}A

\phi=(4j+3k)\times 10^3{\cdot}Ak

Since, k.k=i.i=j.j = 1

So,

\phi=3\times 10^3\times \pi\times (2.66)^2\ k

\phi=6.668\times 10^4\ Nm^2/C

So, the electric flux through a circular area is \phi=6.668\times 10^4\ Nm^2/C.Hence, this is the required solution.

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A skater is using very low friction rollerblades. A friend throws a Frisbee at her, on the straight line along which she is coas
kupik [55]

Answer:

a)  perfectly inelastic,  b)  collision is inelastic,  c)   elastic  

Explanation:

In this exercise, it is asked to identify what type of shock occurs between the skater and the frisbee, for this we must define a system formed by the skater and the fribee, so that the forces during the crash have been internal and the amount of movement is preserved

Initial instant. Before the skater touches the frisbee

    p₀ = M v₁ + m v₂

where M and m are the masses of the skater and frisbee, respectively

for the final moment they give us several possibilities, in all case the moment is conserved

       p₀ = p_{f}

case a)

Final instant. grabs the frisbee and holds it

    p_{f} = (M + m) v '

     p₀ = p_{f}

We can see that this shock is perfectly inelastic, it holds the fressbee

case b)

final instant.

This case is similar to the previous one, but the final speed of fresbee is zero, therefore this collision is inelastic and the kinetic energy is not conserved.

case c)

final instant. Grab the fressbee and resend it

      p_{f} = M v_{1f} + m v_{2f}

this is an elastic Shock since the equivalent of a rebound of the fressbee, the kinetic energy is conserved.

5 0
2 years ago
PLEASE HELP I HAVE NO TIME PLEASEEE DONT SKIP
Triss [41]
2m/s




it has to be 20 charecters just ignore this your answer is up there
7 0
2 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N
Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

r=\sqrt{(r_{1})^2+(r_{2})^2}

Put the value into the formula

r=\sqrt{(9.8)^2+(2.1)^2}

r=10.02\ cm

We need to calculate the force

Using formula of force

F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}

Force F₁₂,

F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}

F_{12}=13.33\ N

F_{21}=-13.33\ N

Force F₂₃,

F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}

We need to calculate the value of third charge

F_{net}=F_{21}+F_{23}

4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}

q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}

q_{3}=7.99\times10^{-7}\ C

q_{3}=0.8\times10^{-6}\ C

Hence, The value of third charge is 0.8μC.

4 0
2 years ago
Does a basketball, baseball, tennis ball, or marble MOST LIKELY have the smallest volume?
Fantom [35]
Marble i think not quite sure doe
4 0
2 years ago
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iVinArrow [24]

Answer:

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Ohm's law : states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance

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