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2 years ago

1) what is the potential energy at the end of the loop?

Physics

1 answer:

Agata [3.3K]2 years ago

8 0

At a roller coaster, energy conversions from potential to kinetic energy and back repeatedly several times across the course of a trip. Kinetic energy is the force that an object possesses as a consequence of its movement.

All moving objects possess kinetic energy, which is defined with the mass and speed of the object. Potential energy is the energy that is deposited in an object due to its position corresponding to some zero point. Gravitational potential energy is highest at the top of the looping point of a roller coaster and smallest at the deepest end.

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A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic fie

Nutka1998 [239]

Answer:

i = 0.5 A

Explanation:

As we know that magnetic flux is given as

$\phi = NBA$

here we know that

N = number of turns

B = magnetic field

A = area of the loop

now we know that rate of change in magnetic flux will induce EMF in the coil

so we have

$EMF = NA\frac{dB}{dt}$

now plug in all values to find induced EMF

$EMF = (20)(50 \times 10^{-4})(\frac{6 - 2}{2})$

$EMF = 0.2 volts$

now by ohm's law we have

$current = \frac{EMF}{Resistance}$

$i = \frac{0.2}{0.40} = 0.5 A$

5 0

3 years ago

Can somebody help me with this quiz please

mezya [45]

Sure what do u need help with

4 0

3 years ago

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$