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Fittoniya [83]
1 year ago
12

Calculate the magnitude of the electric field at the location of given that the square is 6.05 cm on a side.

Physics
1 answer:
Serggg [28]1 year ago
4 0

The term “electric field” refers to the physical field that surrounds electrically charged particles and acts to either attract or repel all other charged particles in the field (also known as an E-field).

It can also refer to the physical field surrounding a system of charged particles. Electric fields are composed of electric charges and time-varying electric currents.

Both electric and magnetic fields are manifestations of the electromagnetic field, one of the four fundamental interactions (sometimes known as forces) of nature.

Electrical technology makes use of electric fields, which are significant in many branches of physics.

For instance, in atomic physics and chemistry, the electric field acts as an attractive force to hold atoms’ atomic nuclei and electrons together. It is also the force that causes atoms to chemically link together to form molecules.

To know more about electric field visit:

brainly.com/question/14811118

#SPJ4

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Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
4 years ago
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