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PilotLPTM [1.2K]
3 years ago
11

I need some help with these two, im really struggling with them.

Physics
2 answers:
MatroZZZ [7]3 years ago
7 0

Answer:a)magnification means the ratio of image distance to object distance i.e

Explanation:

q/p=-2.5

Which means q=-2.5p

Or we can write

Image height (I)over object height (O)

I/O =-2.5

I =-2.5(O)

Which means image is greater by 2.5 times than object and negative sign shows that mirror is convex

Lisa [10]3 years ago
5 0

Answer:

In optics, a magnification refers to the process of enlarging the apparent size of an object, not the physical one, you can call it the virtual size of the object.

Specifically, magnification refers to the number as such, if it's less than one, it refers to a reduction size, which can also be called de-magnification.

For example, a convex lens make things look bigger, there's involved the magnification concept. Another popular example is a microcoscope, which makes small things to look bigger.

So, in this case, a maginification of -2.5 refers to de-magnification, because -2.5 < 1. Basically, the mirror will make objects seem smaller than the real one.

On the other hand, chromatic aberration occurs when a mirror is not able to bring all wavelengths colors into the same plane or focal point, that's why the image seems like distiorted.

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The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.
Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

  • The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, <em>C</em> represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}

The steepness of the slope is therefore;

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100

Where;

\overline{AM} = Half the width of the truck = \dfrac{2.4 \, m}{2} = 1.2 m

\overline{CM} = The elevation of the center of gravity above the ground = 2.2 m

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%

tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}

Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right)  \approx 28.6 ^{\circ}

The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.

Learn more here:

brainly.com/question/20793607

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3 years ago
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A snowball starting at rest rolls down a hill and reaches 5 m/s. If the hill is
Lostsunrise [7]

Answer:

The acceleration of the snowball is 0.3125

Explanation:

The initial speed of the snowball up the hill, u = 0

The speed the snowball reaches, v = 5 m/s

The length of the hill, s = 40 m

The equation of motion of the snowball given the above parameters is therefore;

v² = u² + 2·a·s

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The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .

4 0
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