Why are there shadows?

2 answers:

6 0

Shadows are formed when an opaque object or an object that doesn't allow light to pass through is in the way or infront of etc. a source of light.

lubasha [3.4K]3 years ago

4 0

It is shown when you are covering the light. So like when the sun is shining down at you, you are standing so the stuff behind you is not getting light because you are in its way.
HOPE THIS HELPS!!!!!!! ;-)

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Two charges are located in the xx – yy plane. If ????1=−4.25 nCq1=−4.25 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.0

Sati [7]

Answer:

Ex= -17.1 N/C

Ey = +26.9 N/C

Explanation:

We apply formula of electric field:

Ep=k*q/d²

Ep: Electric field at point ( N/C)

q: Electric charge (C)

k: coulomb constant (N.m²/C²)

d: distance from charge q to point P (m)

In the attached graph we observe the directions of the electric field at P(0,0) due to q1 and q2

Calculation of the field at point P due to the load q₁

E₁=k*q₁/d₁² = 9*10⁹*4.25*10⁻⁹/1.080²= 32.8 N/C : Magnitude of E1

Direction of E₁ :Because the charge q₁ is negative the field enters the charge (+ y)

Calculation of the field at point P due to the load q₂

$d_{2} = \sqrt{1.30^{2}+0.450^{2} }$

d₂=1.375 m

E₂=k*q₂/d₂² = 9*10⁹*3.80*10⁻⁹/ 1.375² = 18.09 N/C Magnitude of E₂

Direction of E₂ :Because the charge q₂ is positive the field leaves the charge in direction of angle β

, then,E₂ tiene componentes x-y en P.

E₂x=-E₂cos β= -18.09*(1.3/1.375)= -17.1 N/C

E₂y=-E₂sin β= -18.09*(0.45/1.375)= -5.9 N/C

Calculation of the electric field at point P located at the origin(0,0)

Ex=E₂x= -17.1 N/C

Ey=E₁y+E₂y =32.8 N/C -5.9 N/C = 26.9 N/C

4 0

3 years ago

What type of gas absorbs and re-emits infrared radiation?

Rashid [163]

Answer: Carbon Dioxide

Explanation:

7 0

3 years ago

Which condition is required for Coulomb's law to hold true?

AleksAgata [21]

The correct answer is:
Point charges must be in a vacuum.

In fact, the usual form for of the Coulomb's law is:
 $F= \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$
where
 $\epsilon_0$ is the permittivity of free space
q1 and q2 are the two charges
q is the separation between the two charges

However, this formula is valid only if the charges are in vacuum. If they are in a material medium, the law is modified as follows:
 $F= \frac{1}{4 \pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{r^2}$
where $\epsilon_r$ is the relative permittivity, which takes into account the dielectric effects of the material.