Answer:
a) The student feel light
b) Nbottom = 758 N
c) N'top= 236 N
d) N'bottom= 1055 N
Explanation:
a) W= 659N , Ntop= 560N
W > Ntop ---> Student feel less weight
b) Top:
∑F= W - Ntop = m.v²/R
m.v²/R = 659N - 560 N = 99 N
Bottom:
∑F= Nbottom- W = m.v²/R
Nbottom= W + m.v²/R = 659N + 99 N = 758N
c) W= 659 N , Ntop= 560 N , v'=2.v
N'top= ?
∑F= W - N'top = m.v'²/R
N'top= W - 4.m.v²/R
N'top = 659 N - 4. 99 N = 263 N
d) N'bottom = ?
∑Fbottom= N'bottom- W = m.v'²/R
N'bottom = W + 4.m.v²/R = 659 N + 4. 99 N = 1055 N
Plug in the corresponding values into y = mx + b
8.18 in for y
1.31 in for m
17.2 in for b
8.18 = 1.31x + 17.2
Now bring 17.2 to the left side by subtracting 17.2 to both sides (what you do on one side you must do to the other). Since 17.2 is being added on the right side, subtraction (the opposite of addition) will cancel it out (make it zero) from the right side and bring it over to the left side.
8.18 - 17.2 = 1.31x
-9.02 = 1.31x
Then divide 1.31 to both sides to isolate x. Since 1.31 is being multiplied by x, division (the opposite of multiplication) will cancel 1.31 out (in this case it will make 1.31 one) from the right side and bring it over to the left side.
-9.02/1.31 = 1.31x/1.31
x ≈ -6.8855
x is roughly -6.89
Hope this helped!
~Just a girl in love with Shawn Mendes
<h2>Answer: True
</h2>
The <u>Doppler effect</u> refers to the change in a wave perceived frequency when the emitter of the waves, and the receiver (or observer in the case of light) move relative to each other.
In other words, it is the variation of the frequency of a wave due to the relative movement of the source of the wave with respect to its receiver.
It should be noted that this effect bears its name in honor of the Austrian physicist <u>Christian Andreas Doppler</u>, who in 1842 proposed the existence of this effect for the case of light in the stars. Another important aspect is that the effect occurs in all waves (including light and sound). However, it is more noticeable to humans with sound waves.
Answer:
This can be translated to:
"find the electrical charge of a body that has 1 million of particles".
First, it will depend on the charge of the particles.
If all the particles have 1 electron more than protons, we will have that the charge of each particle is q = -e = -1.6*10^-19 C
Then the total charge of the body will be:
Q = 1,000,000*-1.6*10^-19 C = -1.6*10^-13 C
If we have the inverse case, where we in each particle we have one more proton than the number of electrons, the total charge will be the opposite of the one of before (because the charge of a proton is equal in magnitude but different in sign than the charge of an electron)
Q = 1.6*10^-13 C
But commonly, we will have a spectrum with the particles, where some of them have a positive charge and some of them will have a negative charge, so we will have a probability of charge that is peaked at Q = 0, this means that, in average, the charge of the particles is canceled by the interaction between them.
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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
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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?
