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1 year ago

Complete the following:

1 answer:

8 0

When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

To find the answer, we have to know about the rules followed by drawing ray-diagram.

<h3>What are the rules obeyed by light rays?</h3>

If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.
The refracted ray becomes parallel to the major axis if the incident ray passes through the focus.
The refracted ray follows the same path if the incident light passes through the center of the curve.

Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

Learn more about refraction by a lens here:

brainly.com/question/13095658

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A plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again. The apparent weight of the g

levacccp [35]

The actual weight of the gas = apparent weight + weight.

The actual weight = $W_{A}$ + W

Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.

If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = $W_{A}$

The gas is squeezed out of the bag to determine its volume by the displacement of water. Since

density = mass / volume

The density of water is 1000 kg/ $m^{2}$

we can get the mass of the gas by making m the subject of the formula.

W = mg

The actual weight of the gas = apparent weight + weight

That is,

The actual weight = $W_{A}$ + W

Learn more about density here: brainly.com/question/406690

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2 years ago

Two children ride side-by-side on a carousel. Their paths are shown in the image below.

Sonbull [250]

Answer:

The child represented by a star on the outside path.

Explanation:

5 0

3 years ago

PLEASE HELP ASAP:

antoniya [11.8K]

True since coulomb's law states that There is electric force between like charges or opposite charges. The negative sign only shows the nature of the force.

<h3>What is the coulombs law ?</h3>

coulombs formula is given by

$F= \dfrac{K\times q_{1} \times q_{2}}{r^{2} }$

Now it states that if two charged particles are separated by the distance r and having same or opposite charge will attract or repel each other.

The intensity of the force depend upon the distance and the nature of the charge.

Hence coulomb's law states that There is electric force between like charges or opposite charges. The negative sign only shows the nature of the force.

To know more about coulomb's law follow

https://brainly.in/question/332179

8 0

2 years ago

Read 2 more answers

Why is it harder to breathe on a

Law Incorporation [45]

I think it’s d but I’m not sure

6 0

2 years ago

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago