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scoray [572]
4 years ago
11

Match each method of transferring electric charge with the correct description

Physics
2 answers:
Mazyrski [523]4 years ago
7 0

Answer:

Friction-1

Conduction-2

Induction-3

Explanation:

Just trust me, bro

babunello [35]4 years ago
4 0

Answer:

FRICTION  OPT 2

INDUCTION OPT 3

CONDUCTION OPT 1

Explanation:

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a light sensor is based on a photodiode that requires a minimum photon energy of 1.30 ev to create mobile electrons. 200nM

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You are 12 miles north of your base camp when you begin walking north at a speed of 2mph. what is your location, relative to you
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3 years ago
The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s.
Aleks [24]

By definition, centripetal acceleration is given by:

a = \frac{v ^ 2}{r}

Where,

v: tangential disk speed

r: disk radius

Substituting values in the given equation we have:

a =\frac{3.3^2}{0.13}\\a = 83.76923077

Rounding the result we have:

a = 83.8 \frac{m}{s^2}

Answer:

The centripetal acceleration of the disc edge in m/s^2 is:

a = 83.8 \frac{m}{s^2}

3 0
4 years ago
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What is rectilinear propagation of light?​
nignag [31]

Answer:

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3 0
3 years ago
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One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren
katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

\sigma_y = \frac{F_y}{A}

where,

\sigma_y =Shear strength

F_y = Shear Force

A_0 =Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}

Resolving for F,

F= 1399.5N

PART B) We need here to apply the shear strength equation, then

\sigma_y = \frac{F_y}{A}

210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}

F_y = 945N

In such a way that the material is more resistant to tensile strength than shear force.

6 0
3 years ago
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