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4 years ago

Match the term to the correct description.

Physics

1 answer:

Lelechka [254]4 years ago

7 0

1. Open circuit

2. Electricity
3. Circuit
4. Series circuit

5. Parallel circuit

6. Resistor

7. Closed circuit

8. switch

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24 A uniform electric field of magnitude 1.1×104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the elect

Tatiana [17]

Answer:

$44,000 Nm^2/C$

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

$\Phi = EA cos \theta$

where:

E is the magnitude of the electric field

A is the area of the surface

$\theta$ is the angle between the direction of the field and of the normal to the surface

In this problem, we have:

$E=1.1\cdot 10^4 N/C$ is the electric field

L = 2.0 m is the side of the sheet, so the area is

$A=L^2=(2.0)^2=4.0 m^2$

$\theta=0^{\circ}$ , since the electric field is perpendicular to the surface

Therefore, the electric flux is

$\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C$

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3 years ago

What two characteristics do electromagnetic waves vary in?

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They size of the wave and the time of a certain wave.

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4 years ago

Which of the following are Electromagnetic Waves?

earnstyle [38]

Answer:

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3 years ago

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

6 0

3 years ago