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svet-max [94.6K]
3 years ago
7

What is the average horizontal component of force exerted on his feet by the ground during acceleration?

Physics
1 answer:
Arte-miy333 [17]3 years ago
7 0
Well you have to think of it like electricity go through your answer closes to that and figure it out
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DEFINE REFRACTION? describe how a ray of light is refracted when it passes through a glass block​
jek_recluse [69]
When a ray passes from air into glass the direction in which the light ray is travelling changes. The light ray appears to bend as it as it passes through the surface of the glass. ... This 'bending of a ray of light' when it passes from one substance into another substance is called refraction.
8 0
2 years ago
Consider electromagnetic waves in free space. What is the wavelength of a wave that has the following frequencies? (a) 4.10 x 10
navik [9.2K]

Explanation:

To find the answer use the equation speed of light=wavelength multiplied by frequency (c=lambda*f) by substituting the value for the frequency the the speed of light

7 0
3 years ago
A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly
Mama L [17]

Answer:

Explanation:

To solve this, we start by using one of the equations of motion. The very first one, in fact

1

V = U + at.

V = 0 + 0.8 * 3.4 = 2.72 m/s.

2.

V = 0 + 0.8 * 4.3 = 3.44 m/s.

3.

d = ½ * 0.8 * 4.3² + 3.44 * 12.9

d = 7.396 + 44.376

d = 51.77 m.

4.

d = 62 - 51.77 = 10.23 m. = Distance

traveled during deceleration.

a = (V² - Vo²) / 2d.

a = (0² - 3.44²) / 20.46

a = -11.8336 / 20.46 = -0.58 m/s²

5.

t = (V - Vo)/a =(0 - 3.44) / -0.58

t = -3.44/-.58 = 5.93 s

= Stop time.

T = 4.3 + 12.9 + 5.93 = 23.13 s. = Total

time the hare was moving.

6.

d = Vo * t + ½ * a * t² = 62 m.

0 + 0.5 * (23.13)² * a = 61

267.5a = 61

a = 61/267.5

a = 0.23 m/s²

7 0
3 years ago
In a light wave, which properties tell you the color of light?
Alenkasestr [34]

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<em><u>Wavel</u></em><em><u>ength</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>distan</u></em><em><u>ce</u></em><em><u> between</u></em><em><u> </u></em><em><u>one</u></em><em><u> </u></em><em><u>crest</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>one</u></em><em><u> </u></em><em><u>through</u></em><em><u> </u></em><em><u>,</u></em><em><u> </u></em><em><u>also</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>dist</u></em><em><u>ance</u></em><em><u> </u></em><em><u>after</u></em><em><u> </u></em><em><u>which</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>wave</u></em><em><u> </u></em><em><u>repe</u></em><em><u>at</u></em><em><u> </u></em><em><u>its</u></em><em><u>elf</u></em><em><u> </u></em><em><u>!</u></em>

<em><u>It's</u></em><em><u> </u></em><em><u>SI</u></em><em><u> </u></em><em><u>unit</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>meter</u></em><em><u> </u></em><em><u>!</u></em><em><u> </u></em>

<em><u>It</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>scalar</u></em><em><u> </u></em><em><u>quan</u></em><em><u>tity</u></em><em><u> </u></em><em><u>!</u></em><em><u>!</u></em><em><u> </u></em>

<em><u>Diff</u></em><em><u>erent</u></em><em><u> </u></em><em><u>Wavelength</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>light</u></em><em><u> </u></em><em><u>have</u></em><em><u> </u></em><em><u>diff</u></em><em><u>erent</u></em><em><u> </u></em><em><u>col</u></em><em><u>or</u></em><em><u> </u></em><em><u>!</u></em><em><u>!</u></em>

<h2>• VIBGYOR </h2>

i.e, Violent , Indigo , Blue , Green , Yellow Orange, and Red along with their shades are the colors which we can see !!

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5 0
3 years ago
An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?
Serjik [45]

The image of the object is 8cm to the left of the lens (D)

<h3></h3>

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

Learn more on focal length here:

brainly.com/question/25779311

#SPJ1

6 0
2 years ago
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