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Allisa [31]
3 years ago
14

Can a fire have a shadow?

Physics
2 answers:
Anton [14]3 years ago
7 0

Answer:

can a fire have a shadow?

Yes it can actually.

edit:Fires can have shadows because they contain hot air and soot, and not because they contain light.

EleoNora [17]3 years ago
3 0

Answer:

No.

Explanation:

A shadow is basically the absence of light. Fire doesn't have a shadow cause fire is itself a source of light, so the wall or obstacle you'd be expecting it's shadow to fall on, would instead be covered by the light from the fire. Hence, fire has no shadow.

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Which of the following is NOT a function of the lens in the eye?
tatyana61 [14]

Answer:

it is a light receptor that generates nerve signals that are sent to the brain

Explanation:

the lens are like the glasses, this means that is used to see things better. You just put them in your eye and that's all it's not connected to the brain

8 0
2 years ago
Brainly the unit of current in which 6,240,000,000,000,000,000 electrons flow past a given point in a circuit in one second is t
BigorU [14]
In electronics, the SI unit for current is Ampere. It is the amount of charge in Coulombs per unit time. It is named after the father of electrodynamics, Andre-Marie Ampere. Also, the current can be easily determined through the Ohm's Law, which states that current is equal to volts divided by the resistance. The answer is letter D.
5 0
3 years ago
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Can someone plz help me DUE TOMORROW
aksik [14]
Hey the answer to the question is
m = 0.40
3 0
3 years ago
Rectangular plate has a voltage of +180V and plate to the first plate) has a 'voltage of -5V. Determie another rectangular plate
3241004551 [841]

For a Rectangular plate has a voltage of +180V and a 'voltage of -5V. , the second plate has the Electric field mathematically given as

E=21.5*10^3v/m

<h3>What is the field strength?</h3>

Generally, the equation for the Electric field   is mathematically given as

E=v/d

Where

v={180-(-5)}v

v=185v

Therefore

E=185/8.6*10^{-3}

E=21.5*10^3v/m

In conclusion, Electric field

E=21.5*10^3v/m

Read more about electric field

brainly.com/question/9383604

4 0
2 years ago
A small sphere of
mixas84 [53]

Answer:

θ = 39.7º

Explanation:

In this exercise we must use Newton's second law for the sphere, at the equilibrium point we write the equations in each exercise; we will assume that plate 1 is on the left

Y Axis

       T_{y} -W = 0

       T_{y} = W

X axis

         -F_{e1}<u> - F_{e2} + Tₓ = 0 </u>

<u> </u>

let's use trigonometry to find the components of the tension, we measure the angle with respect to the vertical

         sin θ = Tₓ / T

         cos θ = T_{y} / t

         Tₓ = T sin θ

         T_{y} = T cos θ

let's use gauss's law to find the electric field of each leaf; We define a Gaussian surface formed by a cylinder, so the component of the field perpendicular to the base of the cylinder is the one with electric flow.

         F = ∫ E. dA = q_{int} / ε₀

  in this case the scalar product is reduced to the algebraic product, the flow is towards both sides of the plate

        F = 2E A = q_{int} / ε₀

let's use the concept of surface charge density

        σ = q_{int} / A

we substitute

        2E A = σ A /ε₀

          E = σ / 2ε₀

this is the field created by each plate. The electric force is

        F_{e} = q E

for plate 1 with σ₁ = -30 10⁻⁶ C / m²

         F_{e1}  = q σ₁ /2ε₀

for plate 2 with s2 = ab 10⁻⁶ C / m², for the calculations a value of this charge density is needed, suppose s2 = 10 10⁻⁶ C / m²

          F_{e2} = q σ₂ /2ε₀

we substitute and write the system of equations

           T cos θ = mg

          - q σ₁ / 2ε₀  - q σ₂ /2ε₀  + T sinθ = 0

we introduce t in the second equations

          - q /2 ε₀  (σ₁ + σ₂) + (mg / cos θ) sin θ = 0

          mg tan θ = q /2ε₀   (σ₁ + σ₂)

          θ = tan -1 (q / 2ε₀ mg (σ₁ + σ₂)

data indicates the mass of 0.25 g = 0.25 10⁻³ kg

give the charge density on plate 2, suppose ab = 10 10⁻⁶ C / m²

let's calculate

         θ = tan⁻¹ (9.0 10⁻¹⁰ (30 + 10) 10⁻⁶ / (2  8.85 10⁻¹² 0.25 10⁻³ 9.8))

         θ = tan⁻¹ 8.3 10⁻¹)

         θ = 39.7º

5 0
3 years ago
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