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3 years ago

1. 3 main components of a circuit

2 answers:

7 0

Voltage, resistance and current are the three components that must be present for a circuit to exist. A circuit will not be able to function without these three components. Voltage is the main electrical source that is present in a circuit. :)

Tomtit [17]3 years ago

3 0

Answer:

resistors
capacitors
inductors
transformers
protection devices - fuses, varistors, etc
transistors
diodes
integrated circuits
processors
SSI and MSI logic
logic and gate arrays and programmable logic
memory
power regulators
voltage references
operational amplifiers
comparators
ASICs, both digital analog and hybrid
converters

connectors
sensors
indicators
switches
mechanical components
The board itself and the conductive traces and special coatings on it.

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The coefficient of friction between a rider and the merry go round is 0.45 and the person is measured to be traveling at 20.0 RP

svlad2 [7]

The person must stand at a radius of 0.99 m

Explanation:

In order for the person to stand on the merry go round, the force of friction acting on the person must provide the centripetal force necessary to keep the person in uniform circular motion.

Therefore, we can write:

$\mu mg = m\omega^2 r$

where:

- the term on the left is the force of friction, and the term on the right is the centripetal force

$\mu = 0.45$ is the coefficient of friction

m is the mass of the person

$g=9.8 m/s^2$ is the acceleration of gravity

$\omega = 20 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=2.09 rad/s$ is the angular velocity

r is the radius of the circular path

Solving the equation for r, we find the radius at which the person must be standing:

$r=\frac{\omega^2}{\mu g}=\frac{(2.09)^2}{(0.45)(9.8)}=0.99 m$

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3 years ago

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generate

soldier1979 [14.2K]

Answer: a) work done = 3946429.5 J

b) work done = 943.22 nutritional calories

Explanation:

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3 years ago

3.00 m3 of a fixed mass of a gas at 150 kPa. Calculate the pressure if the volume is reduced to 1.20 m3 at a constant temperatur

viktelen [127]

Answer:

P₂ = 375 kPa.

Explanation:

Given that,

Initial volume, V₁ = 3 m³

Initial pressure, P₁ = 150 kPa

Final volume, V₂ = 1.2 m³

We need to find the final pressure.

At constant temperature,

$P\propto \dfrac{1}{V}\\\\P_1V_1=P_2V_2\\\\P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{150\ kPa\times 3\ m^3}{1.2\ m^3}\\\\P_2=375\ kPa$

So, the new pressure is 375 kPa.

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3 years ago

What is the potential energy of a 2-kilogram potted plant that is on a 1 meter-high plant stand?

alina1380 [7]

Ep=mgh

In your case it's Ep = 2*9,8*1=19,6 J

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3 years ago

A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (ve

Elanso [62]

The magnitude of the force on the left-hand pole of the thin flexible gold chain of uniform linear density with a mass of 17.1 and, hangs between two 30.0 cm long vertical sticks, which are a distance of 30.0 cm apart will be, 0.167N.

To find the correct answer, we have to know more about the Basic forces that acts upon a body.

<h3>What is force and which are the basic forces that acts upon a body?</h3>

A push or a pull which changes or tends to change the state or rest, or motion of a body is called Force.

Force is a polar vector as it has a point of application.

Positive force represents repulsion and the negative force represented attraction.

There are 3 main forces acting on a body, such as, weight mg, normal reaction N, and the Tension or pulling force.

<h3>How to solve the problem?</h3>

We have given that, the gold chain hangs between the vertical sticks of 30cm and the horizontal distance between then is 30cm.

From the given data, we can find the angle (in the free body diagram, it is given as θ).

$tan\alpha =\frac{30}{30}\\ \alpha =45^0$

From the free body diagram given, we can write the balanced equations of total force along y direction as,

$y-direction\\T_2sin\alpha =mg\\\\T_2=\frac{mg}{sin\alpha } =\frac{17.1*10^{-3}*9.8}{sin45} \\\\T_2=0.236N$

From the free body diagram given, we can write the balanced equations of total force along x direction as,

$x-direction\\T_1-T_2cos\alpha =0\\T_1=T_2cos\alpha \\T_1=0.236*cos45=0.167N$

Thus, we can conclude that, the magnitude of force on the left-hand pole will be 0.167N.

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2 years ago

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