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3 years ago

A type of energy embodied in oscillating electric and magnetic fields is called

Physics

1 answer:

yaroslaw [1]3 years ago

4 0

Answer: Electromagnetic radiation

Explanation:

Electromagnetic radiation is a combination of oscillating electric and magnetic fields, which propagate through space carrying energy from one place to another.

To understand it better:

This radiation is spread thanks to the electromagnetic fields produced by moving electric charges and their sources can be natural or man-made.

It should be noted that the energy of electromagnetic radiation can vary and depending on its frequency it can be useful for various situations.

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Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0

3 years ago

A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is 2.5 m.

Anestetic [448]

Answer:

Explanation:

Diameter of pool = 12 m

radius of pool, r = 6 m

Total height raised, h = 3 + 2.5 = 5.5 m

density of water, d = 1000 kg/m³

Mass of water, m = Volume of water x density

m = πr²h x d

m = 3.14 x 6 x 6 x 5.5 x 1000

m = 113040 kg

Work = m x g x h

W = 113040 x 9.8 x 5.5

W = 6092856 J

7 0

3 years ago

Which conditions are low air pressure systems usually associated with?

Inga [223]

cloudy, wet weather

4 0

3 years ago

Read 2 more answers

You can comfortably hold your fingers close beside a candle flame, but not very close above the flame. why? challenge (optional)

Inessa05 [86]

The candle flame releases hot gases, which directly go in upwards directions. Due to which the air near the flame of the candle is very hot and dense. The particles along with vapour move up. And since the sideways, the air is not very dense and hot, we are able to hold the candle. In anti-gravity region, there will be no density differences and also, the convection process wont occur. So, the candle quickly snuffs off.

6 0

3 years ago

Read 2 more answers

-. A 2kg cart moving to the right at 5m/s collides with an 8kg cart at rest. As a

bulgar [2K]

Answer:

<em>The velocity of the carts after the event is 1 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2+...+m_nv_n$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

$\displaystyle v'=\frac{2*5+8*0}{2+8}=\frac{10}{10}=1$

The velocity of the carts after the event is 1 m/s

3 0

2 years ago