Physics High School

38. Joey set a school record for running the 200 m race. He completed the distance in
20.6s. What was his average speed?

Answers

Answer 1

Answer:

9.7 miles per second

Explanation:

divide 200 by 20.6

Related Questions

A spring extends by 10cm when a mass of 100g is attached to it. What is the spring constant?​

Answers

Answer:

10N/m

Explanation:

Using F=kx

F=mg

k=mg/x

k=0.1*10/0.1 (kg*m/s^2)/m

10N/m


[tex]what \: is \: light \: {?} [/tex]
​

Answers

Light is a form of electromagnetic radiation with a wavelength which can be detected by the human eye. It is a small part of the electromagnetic spectrum and radiation given off by stars like the sun. Animals can also see light. The study of light, known as optics, is an important research area in modern physics.

Answer:

Light is a form of energy which produces the sensation of sight .

hope it is helpful to you

what happens to the piece of plastic?

Answers

Answer:

it reheated and use form another

The Jalapeño pepper is verry spicey almost as spicey as McDonald’s sprite

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?​

Answers

[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}[/tex]

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?

[tex]{\bold{\blue{GIVEN}}}[/tex]

REFRACTIVE INDEX = 1.3

APPARENT DEPTH = 7.7 cm

[tex]{ \bold{\green{To \: Find}}}[/tex]

REAL DEPTH OF THE OBJECT.

[tex]{\red{FORMULA \: \: USED }}[/tex]

[tex]Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth } [/tex]

[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]

Refractive Index = 1.3

Apparent Depth = 7.7 cm

Putting the values in the formula:-

[tex]Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth } \\ \\ 1.3 = \frac{Real \: Depth}{7.7 \: cm} \\ \\ 1.3 \times 7.7 = Real \: Depth \\ \\ 10.01 \: \: cm = Real \: Depth[/tex]

The energy stored by any pair of positive charges is inversely proportional to the distance between them, and directly proportional to their charges. Three identical point charges start at the vertices of an equilateral triangle, and this configuration stores 15 Joules of energy. How much more energy, in Joules, would be stored if one of these charges was moved to the midpoint of the opposite side

Answers

Answer:

 U = 25 J

Explanation:

The energy in a set of charges is given by

          U = [tex]k \sum \frac{q_i}{r_i}[/tex]

in this case we have three charges of equal magnitude

          q = q₁ = q₂ = q₃

with the configuration of an equilateral triangle all distances are worth

          d = a

          U = k ( [tex]\frac{q_1q_2}{ r_1_2 } + \frac{q_1q_3}{r_1_3} + \frac{q_2q_3}{r_2_3}[/tex] )

we substitute

           15 = k q² (3 / a)

            k q² /a = 5

For the second configuration a load is moved to the measured point of the other two

          d₁₃ = a

The distance to charge 2 that is at the midpoint of the other two is

          d₁₂ = d₂₃ = a / 2

           U = k (\frac{q_1q_2}{ r_1_2 } + \frac{q_1q_3}{r_1_3} + \frac{q_2q_3}{r_2_3})

            U = k q² (  [tex]\frac{2}{a} + \frac{1}{a} + \frac{2}{a}[/tex] )

            U = (kq² /a) 5

substituting

            U = 5 5

            U = 25 J

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