PLEASE HELP I'M LOST.

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

LVULAN

3241004551 [841]

For this case we have that by definition, the kinetic energy is given by the following formula:

$k= \frac {1} {2} * m * v ^ 2$

Where:

m: It is the mass

v: It is the velocity

According to the data we have to:

$m = 100 \ kg\\v = 9 \frac {m} {s}$

Substituting the values we have:

$k = \frac {1} {2} * (100) * (9) ^ 2\\k = \frac {1} {2} * (100) * 81\\k = 50 * 81\\k = 4050$

finally, the kinetic energy is $4050 \ J$

Answer:

Option A

7 0

4 years ago

When a solid is in the process of changing into a liquid and extra energy is added to the system, the temperature does not chang

Alekssandra [29.7K]

Answer:

D. The temperature does not change during a phase change because the average kinetic energy does not change. Therefore, the potential energy in the bonds between molecules must change.

Explanation:

When there is a change of state (for example, from solid into a liquid, as in this example), when energy is added to the system, the temperature of the substance does not change.

The reason for this is that the energy supplied is no longer used to increase the average kinetic energy of the particle, but instead it is used to break the bonds between the different particles/molecules. For instance, since in this case the substance is changing from solid to liquid, all the energy supplied during the phase change is used to break the bonds between the molecules of the solid: when the process is done, all the molecules will be free to slide past each other, and the substance has turned completely into a liquid.

The bonds between molecules store potential energy: therefore, this means that the energy supplied during the phase change is not used to change the kinetic energy, but to change the potential energy in the bonds between the molecules.

7 0

3 years ago

Read 2 more answers

What is meant by rusting

Setler79 [48]

The process of formation of a reddish-brown substance on the surface of the iron objects in the presence of flaky moisture and air is called rusting.

3 0

3 years ago

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many lichens are composed of fungi and algae. The fungi get sugars from the algae, and the algae get water, minerals, and protei

Tcecarenko [31]

Answer:

The answer is mutualism because they are both on the receiving and giving ends

6 0

3 years ago