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4 years ago

13

Which object is least likely to allow light to pass through it?

2 answers:

Zarrin [17]4 years ago

8 0

Answer: The correct option is "a tree trunk".

Explanation:

Transparent is a material which allows the light to pass through it.

For example, glass.

An opaque is a material which does not allow light to pass through it.

For example, wood.

Translucent is a material which allows light partially to pass through it.

For example, butter paper.

In the given options, a pair of eyeglasses, a clear water bottle, a car windshield are transparent materials. But a tree trunk is an opaque as it does not light to pass through it.

Therefore, a tree trunk is least likely allow light to pass through it.

jeka57 [31]4 years ago

6 0

a tree truck

is the answer i hope this helps you xD

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Answer:

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Explanation:

Since the equation for the illumination of an object, i.e. the brightness of the light, is <em>inversely proportional to the square of the distance from the light source</em>, the form of the function is:

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3 years ago

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3 years ago

You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determ

Dovator [93]

Answer:

(a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

Explanation:

Given that,

Constant speed = 6.00 m/s

Time = 20.1 sec

Speed = 9.00 m/s

Time = 11.2 sec

We need to write a equation for to travel due north across the river,

Using equation for north

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$6.00^2-c^2=\dfrac{w^2}{(20.1)^2}$

$36-c^2=\dfrac{w^2}{404.01}$ ....(I)

We need to write a equation for to travel due south across the river,

Using equation for south

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$9.00^2-c^2=\dfrac{w^2}{(11.2)^2}$

$81-c^2=\dfrac{w^2}{125.44}$ ....(II)

(a). We need to calculate the wide of the river

Using equation (I) and (II)

$45=\dfrac{w^2}{125.44}-\dfrac{w^2}{404.01}$

$45=w^2(0.00549)$

$w^2=\dfrac{45}{0.00549}$

$w=\sqrt{\dfrac{45}{0.00549}}$

$w=90.5$

We need to calculate the current speed

Using equation (I)

$36-c^2=\dfrac{(90.5)^2}{(20.1)^2}$

$36-c^2=20.27$

$c^2=20.27-36$

$c=\sqrt{15.73}$

$c=3.96\ m/s$

(b). We need to calculate the shortest time

Using formula of time

$t=\dfrac{d}{v}$

$t=\dfrac{90.5}{6}$

$t=15.0\ sec$

We need to calculate the distance

Using formula of distance

$d=vt$

$d=3.96\times15.0$

$d=59.4\ m$

Hence, (a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

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4 years ago