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IgorLugansk [536]
3 years ago
12

What is it called when a surface takes light in without reflecting it

Physics
2 answers:
andrew-mc [135]3 years ago
7 0
Reflection from such a rough surface is called diffuse reflection and appears matte
Travka [436]3 years ago
5 0

Answer: Specular reflection

Explanation:

The law of reflection says that when a ray of light hits a surface, it bounces in a certain way, like a tennis ball thrown against a wall. ... Mirrors, however, don't scatter light in this way. With a smooth surface, light reflects without disturbing the incoming image. This is called specular reflection.

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According to Lenz's law, the induced current in a circuit always flows to oppose the external magnetic field through the circuit
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Answer:

True.

Explanation:

According to Lenz's law, the induced current in a circuit always flows to oppose the external magnetic field through the circuit. This statement is true.

The Faraday's law of induction is given by :

\epsilon=-\dfrac{d\phi}{dt}

Here, negative sign shows that the direction of induced emf is such that opposes the changing current that is its cause.

Hence, the statement is true.

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A well is pumped at Q = 300 m3 /hr in a confined aquifer. The aquifer transmissivity is 25 m2 /hr and the storage coefficient is
Effectus [21]

Answer:

(a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.

(b). The draw-down at a distance 200 m from the well after pumping for 50 hr is 6.707 m.

Explanation:

Given that,

Energy Q=300\ m^3/hr

Transmissivity T = 25\ m^2/hr

Storage coefficient S=2.5\times10^{-4}

Distance r= 200 m

We need to calculate the draw-down at a distance 200 m from the well after pumping for 50 hr

Using formula of draw-down

s= \dfrac{Q}{4\pi T}(-0.5772-ln\dfrac{r^2S}{4tT})

Put the value into the formula

s=\dfrac{300}{4\pi\times25}(-0.5772-ln\dfrac{(200)^2\times2.5\times10^{-4}}{4\times25\times50})

s=5.383\ m

We need to calculate the draw-down at a distance 200 m from the well after pumping for 200 hr

Using formula of draw-down

s= \dfrac{Q}{4\pi T}(-0.5772-ln\dfrac{r^2S}{4tT})

Put the value into the formula

s=\dfrac{300}{4\pi\times25}(-0.5772-ln\dfrac{(200)^2\times2.5\times10^{-4}}{4\times25\times200})

s=6.707\ m

Hence, (a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.

(b). The draw-down at a distance 200 m from the well after pumping for 200 hr is 6.707 m.

3 0
3 years ago
A car is moving at a velocity of 25 ms-1.
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Velocity = 25 + (6x3)= 43 m/s
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