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nexus9112 [7]
3 years ago
9

Helpp pls

Physics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

The intensity of the electric field is

|E|=10654.37 \:N/C

Explanation:

The electric field equation is given by:

|E|=k\frac{q}{d^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge at 0.4100 m from the balloon
  • d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}

|E|=10654.37 \:N/C

I hope it helps you!

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Answer:

(a) \hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

(b) \theta = 85.44^{\circ}

Solution:

As per the question:

Side of the cube, a = 4.4 cm

Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm

Now,

(a) To calculate the unit vector:

\hat{A} = \frac{\vec{A}}{|A|}

\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}

\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}

\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

(b) To calculate the angle between the two vectors say A and A' is given by:

\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta                      

\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})        (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}

Now,

Using equation (1) :

\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})

|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261

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Answer:

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Explanation:

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X-direction     | Y-direction

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Answer:

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