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Leno4ka [110]
3 years ago
8

Sammy is talking to Sally. Position A is a position in front of Sammy's mouth. Position B is a position in front of Sally's ear.

In order for Sally to her Sammy's voice, air particles must _______.
Physics
1 answer:
notka56 [123]3 years ago
3 0

Answer:

Explanation:

To listen the sound, the air particles must vibrates.

As the sound energy produce vibrations in the medium, so the air column exerts pressure, and then the air molecules set into vibrations and we hear sound as the diaphram set into vibrations.

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The cube has a mass of 72.9 g. What substance is this cube composed of?
aalyn [17]
I Need More Details.
But here is somthing i found maybe the same.
brainly.com/question/1354966

8 0
4 years ago
Sound waves reaching the olfactory mucosa do not elicit olfactory perception due to the ____________ not matching the type of st
rodikova [14]

Answer:

Sound waves reaching the olfactory mucosa do not elicit olfactory perception due to the ___MODALITY_________ not matching the type of stimulus of the receptos

Explanation:

Because modality of sensation refers to what is perceived after the stimulus is effected eg the pressure modality is effected when the pressure receptors of the skin are stimulated so in this case sound will not elite olfactory sensation because the modality here is sound rather than smell

3 0
3 years ago
10 turns of wire are closely wound around a pencil as shown in the figure. when measured using a scale as shown, the length of t
Mila [183]

Answer:

a. The thickness of the wire is 2.5 mm.

b. The wire is 0.25 cm thick.

Explanation:

Number of turns of the wire = 10

The length of total turns = 25 mm

a. The thickness of the wire can be determined by;

thickness of the wire = \frac{length of total turns}{number of turns}

                           = \frac{25}{10}

                           = 2.5 mm

Therefore, the wire is 2.5 mm thick.

b. To determine the thickness of the wire in centimetre;

10 mm = 1 cm

So that,

2.5 mm = x

x  = \frac{2.5}{10}

   = 0.25 cm

The wire is 0.25 cm thick.

8 0
3 years ago
How do you change the currents in a circuit
mel-nik [20]
-
Eddy Current Testing

Introduction
Basic Principles
History of ET
Present State of ET

The Physics
Properties of Electricity
Current Flow & Ohm's Law
Induction & Inductance
Self Inductance
Mutual Inductance
Circuits & Phase
Impedance
Depth & Current Density
Phase Lag

Instrumentation
Eddy Current Instruments
Resonant Circuits
Bridges
Impedance Plane
Display - Analog Meter

Probes (Coils)
Probes - Mode of Operation
Probes - Configuration
Probes - Shielding
Coil Design
Impedance Matching

Procedures Issues 
Reference Standards
Signal Filtering

Applications
Surface Breaking Cracks
SBC using Sliding Probes
Tube Inspection 
Conductivity 
Heat Treat Verification
Thickness of Thin Mat'ls
Thickness of Coatings

Advanced Techniques
Scanning
Multi-Frequency Tech.
Swept Frequency Tech.
Pulsed ET Tech.
Background Pulsed ET
Remote Field Tech.

Quizzes

Formulae& Tables
EC Standards & Methods
EC Material Properties
-






Current Flow and Ohm's Law

Ohm's law is the most important, basic law of electricity. It defines the relationship between the three fundamental electrical quantities: current, voltage, and resistance. When a voltage is applied to a circuit containing only resistive elements (i.e. no coils), current flows according to Ohm's Law, which is shown below.

I = V / R 

Where: 

I =

Electrical Current (Amperes)

V =

Voltage (Voltage)

R =

Resistance (Ohms)

    

Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. The formula can be reorganized so that the relationship can easily be seen for all of the three variables.

The Java applet below allows the user to vary each of these three parameters in Ohm's Law and see the effect on the other two parameters. Values may be input into the dialog boxes, or the resistance and voltage may also be varied by moving the arrows in the applet. Current and voltage are shown as they would be displayed on an oscilloscope with the X-axis being time and the Y-axis being the amplitude of the current or voltage. Ohm's Law is valid for both direct current (DC) and alternating current (AC). Note that in AC circuits consisting of purely resistive elements, the current and voltage are always in phase with each other.

Exercise: Use the interactive applet below to investigate the relationship of the variables in Ohm's law. Vary the voltage in the circuit by clicking and dragging the head of the arrow, which is marked with the V. The resistance in the circuit can be increased by dragging the arrow head under the variable resister, which is marked R. Please note that the vertical scale of the oscilloscope screen automatically adjusts to reflect the value of the current.

See what happens to the voltage and current as the resistance in the circuit is increased. What happens if there is not enough resistance in a circuit? If the resistance is increased, what must happen in order to maintain the same level of current flow?


4 0
4 years ago
Slove the following equation for v <br> A=v/t
LenKa [72]

If you meant solve, the answer would be v= A times t

You see, to get rid of an variable on one side, you must do the opposite of what is being done to it, in this case, multiply because division. After this, we must do what we did to one side to the other side. This creates the equation above!

7 0
3 years ago
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